Drawing a circle through 3 non-collinear points











up vote
10
down vote

favorite
1












Does circle through works with 3 points:



documentclass{standalone}
usepackage{tikz}
usetikzlibrary{calc,through}
begin{document}
begin{tikzpicture}
coordinate (A) at (1,1);
coordinate (B) at (2,2);
coordinate (C) at (3,1.5);

node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

foreach i in {A,B,C} {
node[circle,minimum size=1pt,fill=red] at(i) {};
}
end{tikzpicture}
end{document}


I want to draw a circle through A,B and C.



enter image description here










share|improve this question




























    up vote
    10
    down vote

    favorite
    1












    Does circle through works with 3 points:



    documentclass{standalone}
    usepackage{tikz}
    usetikzlibrary{calc,through}
    begin{document}
    begin{tikzpicture}
    coordinate (A) at (1,1);
    coordinate (B) at (2,2);
    coordinate (C) at (3,1.5);

    node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

    foreach i in {A,B,C} {
    node[circle,minimum size=1pt,fill=red] at(i) {};
    }
    end{tikzpicture}
    end{document}


    I want to draw a circle through A,B and C.



    enter image description here










    share|improve this question


























      up vote
      10
      down vote

      favorite
      1









      up vote
      10
      down vote

      favorite
      1






      1





      Does circle through works with 3 points:



      documentclass{standalone}
      usepackage{tikz}
      usetikzlibrary{calc,through}
      begin{document}
      begin{tikzpicture}
      coordinate (A) at (1,1);
      coordinate (B) at (2,2);
      coordinate (C) at (3,1.5);

      node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

      foreach i in {A,B,C} {
      node[circle,minimum size=1pt,fill=red] at(i) {};
      }
      end{tikzpicture}
      end{document}


      I want to draw a circle through A,B and C.



      enter image description here










      share|improve this question















      Does circle through works with 3 points:



      documentclass{standalone}
      usepackage{tikz}
      usetikzlibrary{calc,through}
      begin{document}
      begin{tikzpicture}
      coordinate (A) at (1,1);
      coordinate (B) at (2,2);
      coordinate (C) at (3,1.5);

      node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

      foreach i in {A,B,C} {
      node[circle,minimum size=1pt,fill=red] at(i) {};
      }
      end{tikzpicture}
      end{document}


      I want to draw a circle through A,B and C.



      enter image description here







      tikz-pgf






      share|improve this question















      share|improve this question













      share|improve this question




      share|improve this question








      edited Nov 21 at 19:48









      Artificial Stupidity

      4,8191832




      4,8191832










      asked Nov 21 at 17:02









      lucky1928

      1,0871716




      1,0871716






















          7 Answers
          7






          active

          oldest

          votes

















          up vote
          12
          down vote



          accepted










          The tkz-euclide package has a macro to do this. The manual is written in French.




          1. First, we define the circle with the macro tkzDefCircle.

          2. This macro returns two values that are the center recovered with the macro tkzGetPoint{O}

          3. and the radius that is recovered with the macro tkzGetLength{rayon}.


          Once this is done, we draw the circle with the macro tkzDrawCircle[R](O,rayon pt)



          cercle circonscrit



          documentclass[tikz,border=5mm]{standalone}
          %usepackage{tikz}
          usepackage{tkz-euclide}
          usetikzlibrary{calc,through}
          begin{document}
          begin{tikzpicture}
          coordinate (A) at (1,1);
          coordinate (B) at (2,2);
          coordinate (C) at (3,1.5);

          % node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

          tkzDefCircle[circum](A,B,C)
          tkzGetPoint{O} tkzGetLength{rayon}
          tkzDrawCircle[R](O,rayon pt)

          foreach i in {A,B,C} {
          node[circle,minimum size=1pt,fill=red] at(i) {};
          }
          end{tikzpicture}
          end{document}





          share|improve this answer




























            up vote
            11
            down vote













            A new node style, based on the derivation below plus the information that one can use intersection of p1--p3 and p2--p4, which I learned from AndréC's nice answer



            documentclass[tikz,border=5mm]{standalone}
            usetikzlibrary{calc,through}
            tikzset{circle through 3 points/.style n args={3}{%
            insert path={let p1=($(#1)!0.5!(#2)$),
            p2=($(#1)!0.5!(#3)$),
            p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
            p4=($(#1)!0.5!(#3)!1!90:(#3)$),
            p5=(intersection of p1--p3 and p2--p4)
            in },
            at={(p5)},
            circle through= {(#1)}
            }}

            begin{document}
            begin{tikzpicture}
            coordinate (A) at (1,1);
            coordinate (B) at (2,2);
            coordinate (C) at (3,1.5);
            node[circle through 3 points={A}{B}{C},draw=blue]{};
            foreach i in {A,B,C} {
            node[circle,minimum size=1pt,fill=red] at(i) {};
            }
            end{tikzpicture}
            end{document}


            enter image description here



            Just for fun: an analytic solution based on calc only. (My personal opinion, though, is that this method is more "TikZy", i.e. closer to how the standard TikZ styles work, than the tkz-euclide macros, which are more like pstricks, which I have left behind. However, this is just a personal opinion, and might not be shared by others.)



            documentclass{standalone}
            usepackage{tikz}
            usetikzlibrary{calc}
            tikzset{circle through 3 points/.style n args={3}{%
            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
            }}
            begin{document}
            begin{tikzpicture}
            coordinate (A) at (1,1);
            coordinate (B) at (2,2);
            coordinate (C) at (3,1.5);
            draw[circle through 3 points={A}{B}{C}];
            foreach i in {A,B,C} {
            node[circle,minimum size=1pt,fill=red] at(i) {};
            }
            end{tikzpicture}
            end{document}


            enter image description here



            (Note that n1 is a fraction, and could in principle not be well defined. If you ever encounter this case, just change the ordering, e.g. do draw[circle through 3 points={B}{C}{A}]; or something along those lines.)



            ADDENDUM: Explanation of the analytic formula.



            documentclass[tikz,border=3.14mm]{standalone}
            usepackage{tikz}
            usetikzlibrary{calc}
            tikzset{circle through 3 points/.style n args={3}{%
            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
            }}
            begin{document}
            foreach X in {1,...,5}
            {begin{tikzpicture}[font=sffamily]
            path[use as bounding box] (-1,-4) rectangle (6,4);
            coordinate (A) at (1,1);
            coordinate (B) at (2,2);
            coordinate (C) at (3,1.5);
            foreach i in {A,B,C} {
            node[circle,minimum size=1pt,fill=red] at(i) {};
            }
            ifnumX=1
            node[anchor=north,text width=7cm] (start) at (2.5,0){Starting point: 3 points.};
            foreach Y in {A,B,C}
            {draw[-latex] (start) to[out=90,in=-90] (Y) node[above=2pt]{Y}; }
            fi
            ifnumX=2
            coordinate (auxAB) at ($ (A)!.5!(B) $);
            coordinate (auxBC) at ($ (B)!.5!(C) $);
            draw (A) -- (B) -- (C);
            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $) coordinate(aux1);
            draw ($ (auxBC)!1.2cm!90:(B) $) coordinate(aux2) -- ($ (auxBC)!1.2cm!-90:(B) $);
            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
            where the lines that run through and are orthogonal to the edges intersect.};
            draw[-latex] (int) to[out=45,in=-90] (aux1);
            draw[-latex] (int) to[out=135,in=-90] (aux2);
            fi
            ifnumX=3
            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
            foreach Y in {auxAB,auxBC}
            {fill (Y) circle (1pt);}
            draw (A) -- (B) -- (C);
            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
            node[anchor=north,text width=7cm] (int) at (2.5,0){Call the points in the
            middle $P_2$ and $P_4$, and the differences $P_1=A-B$ and $P_3=B-C$. Then the
            orthogonal lines will fulfill
            [gamma_1(alpha)~=~left(begin{array}{c}
            x_2+alpha,y_1\
            y_2-alpha,x_1\
            end{array}right) ]
            and
            [gamma_2(beta)~=~left(begin{array}{c}
            x_4+beta,y_3\
            y_4-beta,x_3\
            end{array}right);. ]
            };
            fi
            ifnumX=4
            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
            foreach Y in {auxAB,auxBC}
            {fill (Y) circle (1pt);}
            draw (A) -- (B) -- (C);
            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
            then simply determined by
            [gamma_1(alpha)~=~gamma_2(beta);, ]
            which has the solution
            [
            alpha~=~frac{-(x_2cdot x_3) + x_3cdot x_4 + y_3cdot (y_4-y_2 )}{x_3cdot y_1 - x_1cdot y_3};.
            ]
            This is texttt{textbackslash n1} in the Tiemph{k}Z style texttt{circle through 3 points}.
            };
            fi
            ifnumX=5
            draw[circle through 3 points={A}{B}{C}];
            node[anchor=north,text width=7cm] (int) at (2.5,-0.1){Once we have the center,
            determining the radius (texttt{textbackslash n2}) is trivial, and we can draw
            the circle with a simple texttt{insert path}.};
            fi
            end{tikzpicture}}
            end{document}


            enter image description here






            share|improve this answer























            • (+1) Give the analytical expression also, if possible :)
              – nidhin
              Nov 21 at 19:16










            • @nidhin It is in the code, isn't it?
              – marmot
              Nov 21 at 19:20










            • Yes it is there. I meant outside the code. Mathematical expression.
              – nidhin
              Nov 21 at 19:25










            • @nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create.
              – marmot
              Nov 21 at 19:54










            • Wow. That was more than expected. :)
              – nidhin
              Nov 21 at 20:03


















            up vote
            7
            down vote













            You can use tkz-euclide like this:



            documentclass{standalone}
            usepackage{tikz}
            usepackage{tkz-euclide}
            usetikzlibrary{calc,through}
            begin{document}
            begin{tikzpicture}
            coordinate (A) at (1,1);
            coordinate (B) at (2,2);
            coordinate (C) at (3,1.5);

            node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

            foreach i in {A,B,C} {
            node[circle,minimum size=1pt,fill=red] at(i) {};
            }
            tkzCircumCenter(A,B,C)tkzGetPoint{O}
            tkzDrawCircle(O,A)
            end{tikzpicture}
            end{document}


            (Modified from https://tex.stackexchange.com/a/16024/8650)
            Circles



            If you chose to use tkz-euclide, then you should consider to do all of your drawing with it - depending on what it is - you can e.g. define your points with tkzDefPoint(x,y).






            share|improve this answer




























              up vote
              7
              down vote













              Just for comparison purpose.



              documentclass[pstricks]{standalone}  
              usepackage{pst-eucl}
              begin{document}
              foreach i in {1.0,1.2,...,4.0}{
              begin{pspicture}(-5,-5)(5,5)
              pstTriangle(4;30){A}(4;90){B}(i;-45){C}
              pstCircleABC{A}{B}{C}{O}
              end{pspicture}}
              end{document}


              enter image description here






              share|improve this answer




























                up vote
                6
                down vote













                The code



                node [draw] at (1,1) [circle through={(A)}] {};


                draw a circle whose center is at (1,1) and passes through A. In this case the center of the Circumscribed circle has to be calculated before using through.



                I just used the Straightedge and compass construction to calculate the center and then drew circle. The basic idea is that all the perpendicular bisectors of the edges of a triangle meet at the same point: the circumcenter.



                enter image description here



                documentclass{standalone}
                usepackage{tikz}
                usetikzlibrary{through,intersections}
                begin{document}
                begin{tikzpicture}
                coordinate (A) at (1,1);
                coordinate (B) at (2,2);
                coordinate (C) at (3,1.5);
                path[name path=c1] (A) circle[radius=5cm];
                path[name path=c2] (B) circle[radius=5cm];
                path[name path=c3] (C) circle[radius=5cm];
                path[name intersections={of = c1 and c2}];
                path[name path=o1] (intersection-1)--(intersection-2);
                path[name intersections={of = c2 and c3}];
                path[name path=o2] (intersection-1)--(intersection-2);
                path[name intersections={of = o1 and o2}];
                node[draw,line width=2pt] at (intersection-1) [circle through={(A)}]{};

                foreach i in {A,B,C} {
                node[circle,minimum size=1pt,fill=red] at(i) {};
                }
                end{tikzpicture}
                end{document}





                share|improve this answer






























                  up vote
                  5
                  down vote













                  Just for fun, another solution inspired by the @marmot solution that calculates the intersection of two defined perpendicular bisector with the calc library.



                  cercle



                  documentclass[tikz,border=5mm]{standalone}
                  usetikzlibrary{calc,through}
                  tikzset{circle through 3 points/.style n args={3}{%
                  insert path={let p1=($(#1)!0.5!(#2)$),
                  p2=($(#1)!0.5!(#3)$),
                  p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                  p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                  p5=(intersection of p1--p3 and p2--p4)
                  in
                  node at (p5) [draw,line width=2pt,circle through= {(#1)}]{}}
                  }}

                  begin{document}
                  begin{tikzpicture}
                  coordinate (A) at (1,1);
                  coordinate (B) at (2,2);
                  coordinate (C) at (3,1.5);
                  draw[circle through 3 points={A}{B}{C}];
                  foreach i in {A,B,C} {
                  node[circle,minimum size=1pt,fill=red] at(i) {};
                  }
                  end{tikzpicture}
                  end{document}





                  share|improve this answer





















                  • Nice! I didn't know that one could use p1 etc. in intersection of p1--p3 and p2--p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what path usually does.
                    – marmot
                    Nov 21 at 21:11










                  • @marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
                    – AndréC
                    Nov 21 at 21:56












                  • I really could only do that after I learned the intersection of p1--p3 and p2--p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
                    – marmot
                    Nov 21 at 22:03










                  • @marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
                    – AndréC
                    Nov 21 at 22:07










                  • I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
                    – marmot
                    Nov 21 at 22:11


















                  up vote
                  5
                  down vote













                  Just for fun with @AndréC's answer:



                  documentclass[tikz,border=5mm]{standalone}
                  usetikzlibrary{calc,through}
                  begin{document}
                  begin{tikzpicture}
                  coordinate (A) at (1,1);
                  coordinate (B) at (2,2);
                  coordinate (C) at (3,1.5);

                  draw let p1=($(A)!0.5!(B)$),
                  p2=($(A)!0.5!(C)$),
                  p3=($(p1)!2!-90:(B)$),
                  p4=($(p1)!2!90:(B)$),
                  p5=($(p2)!2!-90:(C)$),
                  p6=($(p2)!2!90:(C)$),
                  p7=(intersection of p3--p4 and p5--p6)
                  in
                  (A) -- (B)
                  (A) -- (C)
                  (p3) -- (p4)
                  (p5) -- (p6)
                  foreach j in {1,...,7} {
                  node[circle,minimum size=2pt,fill=red,inner sep=0,label=j] at(pj) {}
                  }
                  node[draw,line width=1pt,circle through= {(A)}] at (p7) {};

                  foreach i in {A,B,C} {
                  node[circle,minimum size=5pt,fill=red,inner sep=0,label=i] at(i) {};
                  }
                  end{tikzpicture}
                  end{document}


                  enter image description here






                  share|improve this answer























                    Your Answer








                    StackExchange.ready(function() {
                    var channelOptions = {
                    tags: "".split(" "),
                    id: "85"
                    };
                    initTagRenderer("".split(" "), "".split(" "), channelOptions);

                    StackExchange.using("externalEditor", function() {
                    // Have to fire editor after snippets, if snippets enabled
                    if (StackExchange.settings.snippets.snippetsEnabled) {
                    StackExchange.using("snippets", function() {
                    createEditor();
                    });
                    }
                    else {
                    createEditor();
                    }
                    });

                    function createEditor() {
                    StackExchange.prepareEditor({
                    heartbeatType: 'answer',
                    convertImagesToLinks: false,
                    noModals: true,
                    showLowRepImageUploadWarning: true,
                    reputationToPostImages: null,
                    bindNavPrevention: true,
                    postfix: "",
                    imageUploader: {
                    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
                    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
                    allowUrls: true
                    },
                    onDemand: true,
                    discardSelector: ".discard-answer"
                    ,immediatelyShowMarkdownHelp:true
                    });


                    }
                    });














                     

                    draft saved


                    draft discarded


















                    StackExchange.ready(
                    function () {
                    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f461161%2fdrawing-a-circle-through-3-non-collinear-points%23new-answer', 'question_page');
                    }
                    );

                    Post as a guest















                    Required, but never shown

























                    7 Answers
                    7






                    active

                    oldest

                    votes








                    7 Answers
                    7






                    active

                    oldest

                    votes









                    active

                    oldest

                    votes






                    active

                    oldest

                    votes








                    up vote
                    12
                    down vote



                    accepted










                    The tkz-euclide package has a macro to do this. The manual is written in French.




                    1. First, we define the circle with the macro tkzDefCircle.

                    2. This macro returns two values that are the center recovered with the macro tkzGetPoint{O}

                    3. and the radius that is recovered with the macro tkzGetLength{rayon}.


                    Once this is done, we draw the circle with the macro tkzDrawCircle[R](O,rayon pt)



                    cercle circonscrit



                    documentclass[tikz,border=5mm]{standalone}
                    %usepackage{tikz}
                    usepackage{tkz-euclide}
                    usetikzlibrary{calc,through}
                    begin{document}
                    begin{tikzpicture}
                    coordinate (A) at (1,1);
                    coordinate (B) at (2,2);
                    coordinate (C) at (3,1.5);

                    % node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                    tkzDefCircle[circum](A,B,C)
                    tkzGetPoint{O} tkzGetLength{rayon}
                    tkzDrawCircle[R](O,rayon pt)

                    foreach i in {A,B,C} {
                    node[circle,minimum size=1pt,fill=red] at(i) {};
                    }
                    end{tikzpicture}
                    end{document}





                    share|improve this answer

























                      up vote
                      12
                      down vote



                      accepted










                      The tkz-euclide package has a macro to do this. The manual is written in French.




                      1. First, we define the circle with the macro tkzDefCircle.

                      2. This macro returns two values that are the center recovered with the macro tkzGetPoint{O}

                      3. and the radius that is recovered with the macro tkzGetLength{rayon}.


                      Once this is done, we draw the circle with the macro tkzDrawCircle[R](O,rayon pt)



                      cercle circonscrit



                      documentclass[tikz,border=5mm]{standalone}
                      %usepackage{tikz}
                      usepackage{tkz-euclide}
                      usetikzlibrary{calc,through}
                      begin{document}
                      begin{tikzpicture}
                      coordinate (A) at (1,1);
                      coordinate (B) at (2,2);
                      coordinate (C) at (3,1.5);

                      % node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                      tkzDefCircle[circum](A,B,C)
                      tkzGetPoint{O} tkzGetLength{rayon}
                      tkzDrawCircle[R](O,rayon pt)

                      foreach i in {A,B,C} {
                      node[circle,minimum size=1pt,fill=red] at(i) {};
                      }
                      end{tikzpicture}
                      end{document}





                      share|improve this answer























                        up vote
                        12
                        down vote



                        accepted







                        up vote
                        12
                        down vote



                        accepted






                        The tkz-euclide package has a macro to do this. The manual is written in French.




                        1. First, we define the circle with the macro tkzDefCircle.

                        2. This macro returns two values that are the center recovered with the macro tkzGetPoint{O}

                        3. and the radius that is recovered with the macro tkzGetLength{rayon}.


                        Once this is done, we draw the circle with the macro tkzDrawCircle[R](O,rayon pt)



                        cercle circonscrit



                        documentclass[tikz,border=5mm]{standalone}
                        %usepackage{tikz}
                        usepackage{tkz-euclide}
                        usetikzlibrary{calc,through}
                        begin{document}
                        begin{tikzpicture}
                        coordinate (A) at (1,1);
                        coordinate (B) at (2,2);
                        coordinate (C) at (3,1.5);

                        % node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                        tkzDefCircle[circum](A,B,C)
                        tkzGetPoint{O} tkzGetLength{rayon}
                        tkzDrawCircle[R](O,rayon pt)

                        foreach i in {A,B,C} {
                        node[circle,minimum size=1pt,fill=red] at(i) {};
                        }
                        end{tikzpicture}
                        end{document}





                        share|improve this answer












                        The tkz-euclide package has a macro to do this. The manual is written in French.




                        1. First, we define the circle with the macro tkzDefCircle.

                        2. This macro returns two values that are the center recovered with the macro tkzGetPoint{O}

                        3. and the radius that is recovered with the macro tkzGetLength{rayon}.


                        Once this is done, we draw the circle with the macro tkzDrawCircle[R](O,rayon pt)



                        cercle circonscrit



                        documentclass[tikz,border=5mm]{standalone}
                        %usepackage{tikz}
                        usepackage{tkz-euclide}
                        usetikzlibrary{calc,through}
                        begin{document}
                        begin{tikzpicture}
                        coordinate (A) at (1,1);
                        coordinate (B) at (2,2);
                        coordinate (C) at (3,1.5);

                        % node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                        tkzDefCircle[circum](A,B,C)
                        tkzGetPoint{O} tkzGetLength{rayon}
                        tkzDrawCircle[R](O,rayon pt)

                        foreach i in {A,B,C} {
                        node[circle,minimum size=1pt,fill=red] at(i) {};
                        }
                        end{tikzpicture}
                        end{document}






                        share|improve this answer












                        share|improve this answer



                        share|improve this answer










                        answered Nov 21 at 17:41









                        AndréC

                        6,24711140




                        6,24711140






















                            up vote
                            11
                            down vote













                            A new node style, based on the derivation below plus the information that one can use intersection of p1--p3 and p2--p4, which I learned from AndréC's nice answer



                            documentclass[tikz,border=5mm]{standalone}
                            usetikzlibrary{calc,through}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)!0.5!(#2)$),
                            p2=($(#1)!0.5!(#3)$),
                            p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                            p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                            p5=(intersection of p1--p3 and p2--p4)
                            in },
                            at={(p5)},
                            circle through= {(#1)}
                            }}

                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            node[circle through 3 points={A}{B}{C},draw=blue]{};
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            Just for fun: an analytic solution based on calc only. (My personal opinion, though, is that this method is more "TikZy", i.e. closer to how the standard TikZ styles work, than the tkz-euclide macros, which are more like pstricks, which I have left behind. However, this is just a personal opinion, and might not be shared by others.)



                            documentclass{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            draw[circle through 3 points={A}{B}{C}];
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            (Note that n1 is a fraction, and could in principle not be well defined. If you ever encounter this case, just change the ordering, e.g. do draw[circle through 3 points={B}{C}{A}]; or something along those lines.)



                            ADDENDUM: Explanation of the analytic formula.



                            documentclass[tikz,border=3.14mm]{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            foreach X in {1,...,5}
                            {begin{tikzpicture}[font=sffamily]
                            path[use as bounding box] (-1,-4) rectangle (6,4);
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            ifnumX=1
                            node[anchor=north,text width=7cm] (start) at (2.5,0){Starting point: 3 points.};
                            foreach Y in {A,B,C}
                            {draw[-latex] (start) to[out=90,in=-90] (Y) node[above=2pt]{Y}; }
                            fi
                            ifnumX=2
                            coordinate (auxAB) at ($ (A)!.5!(B) $);
                            coordinate (auxBC) at ($ (B)!.5!(C) $);
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $) coordinate(aux1);
                            draw ($ (auxBC)!1.2cm!90:(B) $) coordinate(aux2) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            where the lines that run through and are orthogonal to the edges intersect.};
                            draw[-latex] (int) to[out=45,in=-90] (aux1);
                            draw[-latex] (int) to[out=135,in=-90] (aux2);
                            fi
                            ifnumX=3
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){Call the points in the
                            middle $P_2$ and $P_4$, and the differences $P_1=A-B$ and $P_3=B-C$. Then the
                            orthogonal lines will fulfill
                            [gamma_1(alpha)~=~left(begin{array}{c}
                            x_2+alpha,y_1\
                            y_2-alpha,x_1\
                            end{array}right) ]
                            and
                            [gamma_2(beta)~=~left(begin{array}{c}
                            x_4+beta,y_3\
                            y_4-beta,x_3\
                            end{array}right);. ]
                            };
                            fi
                            ifnumX=4
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            then simply determined by
                            [gamma_1(alpha)~=~gamma_2(beta);, ]
                            which has the solution
                            [
                            alpha~=~frac{-(x_2cdot x_3) + x_3cdot x_4 + y_3cdot (y_4-y_2 )}{x_3cdot y_1 - x_1cdot y_3};.
                            ]
                            This is texttt{textbackslash n1} in the Tiemph{k}Z style texttt{circle through 3 points}.
                            };
                            fi
                            ifnumX=5
                            draw[circle through 3 points={A}{B}{C}];
                            node[anchor=north,text width=7cm] (int) at (2.5,-0.1){Once we have the center,
                            determining the radius (texttt{textbackslash n2}) is trivial, and we can draw
                            the circle with a simple texttt{insert path}.};
                            fi
                            end{tikzpicture}}
                            end{document}


                            enter image description here






                            share|improve this answer























                            • (+1) Give the analytical expression also, if possible :)
                              – nidhin
                              Nov 21 at 19:16










                            • @nidhin It is in the code, isn't it?
                              – marmot
                              Nov 21 at 19:20










                            • Yes it is there. I meant outside the code. Mathematical expression.
                              – nidhin
                              Nov 21 at 19:25










                            • @nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create.
                              – marmot
                              Nov 21 at 19:54










                            • Wow. That was more than expected. :)
                              – nidhin
                              Nov 21 at 20:03















                            up vote
                            11
                            down vote













                            A new node style, based on the derivation below plus the information that one can use intersection of p1--p3 and p2--p4, which I learned from AndréC's nice answer



                            documentclass[tikz,border=5mm]{standalone}
                            usetikzlibrary{calc,through}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)!0.5!(#2)$),
                            p2=($(#1)!0.5!(#3)$),
                            p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                            p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                            p5=(intersection of p1--p3 and p2--p4)
                            in },
                            at={(p5)},
                            circle through= {(#1)}
                            }}

                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            node[circle through 3 points={A}{B}{C},draw=blue]{};
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            Just for fun: an analytic solution based on calc only. (My personal opinion, though, is that this method is more "TikZy", i.e. closer to how the standard TikZ styles work, than the tkz-euclide macros, which are more like pstricks, which I have left behind. However, this is just a personal opinion, and might not be shared by others.)



                            documentclass{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            draw[circle through 3 points={A}{B}{C}];
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            (Note that n1 is a fraction, and could in principle not be well defined. If you ever encounter this case, just change the ordering, e.g. do draw[circle through 3 points={B}{C}{A}]; or something along those lines.)



                            ADDENDUM: Explanation of the analytic formula.



                            documentclass[tikz,border=3.14mm]{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            foreach X in {1,...,5}
                            {begin{tikzpicture}[font=sffamily]
                            path[use as bounding box] (-1,-4) rectangle (6,4);
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            ifnumX=1
                            node[anchor=north,text width=7cm] (start) at (2.5,0){Starting point: 3 points.};
                            foreach Y in {A,B,C}
                            {draw[-latex] (start) to[out=90,in=-90] (Y) node[above=2pt]{Y}; }
                            fi
                            ifnumX=2
                            coordinate (auxAB) at ($ (A)!.5!(B) $);
                            coordinate (auxBC) at ($ (B)!.5!(C) $);
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $) coordinate(aux1);
                            draw ($ (auxBC)!1.2cm!90:(B) $) coordinate(aux2) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            where the lines that run through and are orthogonal to the edges intersect.};
                            draw[-latex] (int) to[out=45,in=-90] (aux1);
                            draw[-latex] (int) to[out=135,in=-90] (aux2);
                            fi
                            ifnumX=3
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){Call the points in the
                            middle $P_2$ and $P_4$, and the differences $P_1=A-B$ and $P_3=B-C$. Then the
                            orthogonal lines will fulfill
                            [gamma_1(alpha)~=~left(begin{array}{c}
                            x_2+alpha,y_1\
                            y_2-alpha,x_1\
                            end{array}right) ]
                            and
                            [gamma_2(beta)~=~left(begin{array}{c}
                            x_4+beta,y_3\
                            y_4-beta,x_3\
                            end{array}right);. ]
                            };
                            fi
                            ifnumX=4
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            then simply determined by
                            [gamma_1(alpha)~=~gamma_2(beta);, ]
                            which has the solution
                            [
                            alpha~=~frac{-(x_2cdot x_3) + x_3cdot x_4 + y_3cdot (y_4-y_2 )}{x_3cdot y_1 - x_1cdot y_3};.
                            ]
                            This is texttt{textbackslash n1} in the Tiemph{k}Z style texttt{circle through 3 points}.
                            };
                            fi
                            ifnumX=5
                            draw[circle through 3 points={A}{B}{C}];
                            node[anchor=north,text width=7cm] (int) at (2.5,-0.1){Once we have the center,
                            determining the radius (texttt{textbackslash n2}) is trivial, and we can draw
                            the circle with a simple texttt{insert path}.};
                            fi
                            end{tikzpicture}}
                            end{document}


                            enter image description here






                            share|improve this answer























                            • (+1) Give the analytical expression also, if possible :)
                              – nidhin
                              Nov 21 at 19:16










                            • @nidhin It is in the code, isn't it?
                              – marmot
                              Nov 21 at 19:20










                            • Yes it is there. I meant outside the code. Mathematical expression.
                              – nidhin
                              Nov 21 at 19:25










                            • @nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create.
                              – marmot
                              Nov 21 at 19:54










                            • Wow. That was more than expected. :)
                              – nidhin
                              Nov 21 at 20:03













                            up vote
                            11
                            down vote










                            up vote
                            11
                            down vote









                            A new node style, based on the derivation below plus the information that one can use intersection of p1--p3 and p2--p4, which I learned from AndréC's nice answer



                            documentclass[tikz,border=5mm]{standalone}
                            usetikzlibrary{calc,through}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)!0.5!(#2)$),
                            p2=($(#1)!0.5!(#3)$),
                            p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                            p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                            p5=(intersection of p1--p3 and p2--p4)
                            in },
                            at={(p5)},
                            circle through= {(#1)}
                            }}

                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            node[circle through 3 points={A}{B}{C},draw=blue]{};
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            Just for fun: an analytic solution based on calc only. (My personal opinion, though, is that this method is more "TikZy", i.e. closer to how the standard TikZ styles work, than the tkz-euclide macros, which are more like pstricks, which I have left behind. However, this is just a personal opinion, and might not be shared by others.)



                            documentclass{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            draw[circle through 3 points={A}{B}{C}];
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            (Note that n1 is a fraction, and could in principle not be well defined. If you ever encounter this case, just change the ordering, e.g. do draw[circle through 3 points={B}{C}{A}]; or something along those lines.)



                            ADDENDUM: Explanation of the analytic formula.



                            documentclass[tikz,border=3.14mm]{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            foreach X in {1,...,5}
                            {begin{tikzpicture}[font=sffamily]
                            path[use as bounding box] (-1,-4) rectangle (6,4);
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            ifnumX=1
                            node[anchor=north,text width=7cm] (start) at (2.5,0){Starting point: 3 points.};
                            foreach Y in {A,B,C}
                            {draw[-latex] (start) to[out=90,in=-90] (Y) node[above=2pt]{Y}; }
                            fi
                            ifnumX=2
                            coordinate (auxAB) at ($ (A)!.5!(B) $);
                            coordinate (auxBC) at ($ (B)!.5!(C) $);
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $) coordinate(aux1);
                            draw ($ (auxBC)!1.2cm!90:(B) $) coordinate(aux2) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            where the lines that run through and are orthogonal to the edges intersect.};
                            draw[-latex] (int) to[out=45,in=-90] (aux1);
                            draw[-latex] (int) to[out=135,in=-90] (aux2);
                            fi
                            ifnumX=3
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){Call the points in the
                            middle $P_2$ and $P_4$, and the differences $P_1=A-B$ and $P_3=B-C$. Then the
                            orthogonal lines will fulfill
                            [gamma_1(alpha)~=~left(begin{array}{c}
                            x_2+alpha,y_1\
                            y_2-alpha,x_1\
                            end{array}right) ]
                            and
                            [gamma_2(beta)~=~left(begin{array}{c}
                            x_4+beta,y_3\
                            y_4-beta,x_3\
                            end{array}right);. ]
                            };
                            fi
                            ifnumX=4
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            then simply determined by
                            [gamma_1(alpha)~=~gamma_2(beta);, ]
                            which has the solution
                            [
                            alpha~=~frac{-(x_2cdot x_3) + x_3cdot x_4 + y_3cdot (y_4-y_2 )}{x_3cdot y_1 - x_1cdot y_3};.
                            ]
                            This is texttt{textbackslash n1} in the Tiemph{k}Z style texttt{circle through 3 points}.
                            };
                            fi
                            ifnumX=5
                            draw[circle through 3 points={A}{B}{C}];
                            node[anchor=north,text width=7cm] (int) at (2.5,-0.1){Once we have the center,
                            determining the radius (texttt{textbackslash n2}) is trivial, and we can draw
                            the circle with a simple texttt{insert path}.};
                            fi
                            end{tikzpicture}}
                            end{document}


                            enter image description here






                            share|improve this answer














                            A new node style, based on the derivation below plus the information that one can use intersection of p1--p3 and p2--p4, which I learned from AndréC's nice answer



                            documentclass[tikz,border=5mm]{standalone}
                            usetikzlibrary{calc,through}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)!0.5!(#2)$),
                            p2=($(#1)!0.5!(#3)$),
                            p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                            p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                            p5=(intersection of p1--p3 and p2--p4)
                            in },
                            at={(p5)},
                            circle through= {(#1)}
                            }}

                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            node[circle through 3 points={A}{B}{C},draw=blue]{};
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            Just for fun: an analytic solution based on calc only. (My personal opinion, though, is that this method is more "TikZy", i.e. closer to how the standard TikZ styles work, than the tkz-euclide macros, which are more like pstricks, which I have left behind. However, this is just a personal opinion, and might not be shared by others.)



                            documentclass{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            draw[circle through 3 points={A}{B}{C}];
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            end{tikzpicture}
                            end{document}


                            enter image description here



                            (Note that n1 is a fraction, and could in principle not be well defined. If you ever encounter this case, just change the ordering, e.g. do draw[circle through 3 points={B}{C}{A}]; or something along those lines.)



                            ADDENDUM: Explanation of the analytic formula.



                            documentclass[tikz,border=3.14mm]{standalone}
                            usepackage{tikz}
                            usetikzlibrary{calc}
                            tikzset{circle through 3 points/.style n args={3}{%
                            insert path={let p1=($(#1)-(#2)$),p2=($(#1)!0.5!(#2)$),
                            p3=($(#1)-(#3)$),p4=($(#1)!0.5!(#3)$),p5=(#1),n1={(-(x2*x3) + x3*x4 + y3*(-y2 +
                            y4))/(x3*y1 - x1*y3)},n2={veclen(x5-x2-n1*y1,y5-y2+n1*x1)} in
                            ({x2+n1*y1},{y2-n1*x1}) circle (n2)}
                            }}
                            begin{document}
                            foreach X in {1,...,5}
                            {begin{tikzpicture}[font=sffamily]
                            path[use as bounding box] (-1,-4) rectangle (6,4);
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);
                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            ifnumX=1
                            node[anchor=north,text width=7cm] (start) at (2.5,0){Starting point: 3 points.};
                            foreach Y in {A,B,C}
                            {draw[-latex] (start) to[out=90,in=-90] (Y) node[above=2pt]{Y}; }
                            fi
                            ifnumX=2
                            coordinate (auxAB) at ($ (A)!.5!(B) $);
                            coordinate (auxBC) at ($ (B)!.5!(C) $);
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $) coordinate(aux1);
                            draw ($ (auxBC)!1.2cm!90:(B) $) coordinate(aux2) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            where the lines that run through and are orthogonal to the edges intersect.};
                            draw[-latex] (int) to[out=45,in=-90] (aux1);
                            draw[-latex] (int) to[out=135,in=-90] (aux2);
                            fi
                            ifnumX=3
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){Call the points in the
                            middle $P_2$ and $P_4$, and the differences $P_1=A-B$ and $P_3=B-C$. Then the
                            orthogonal lines will fulfill
                            [gamma_1(alpha)~=~left(begin{array}{c}
                            x_2+alpha,y_1\
                            y_2-alpha,x_1\
                            end{array}right) ]
                            and
                            [gamma_2(beta)~=~left(begin{array}{c}
                            x_4+beta,y_3\
                            y_4-beta,x_3\
                            end{array}right);. ]
                            };
                            fi
                            ifnumX=4
                            coordinate[label=below:$P_2$] (auxAB) at ($ (A)!.5!(B) $);
                            coordinate[label=below:$P_4$] (auxBC) at ($ (B)!.5!(C) $);
                            foreach Y in {auxAB,auxBC}
                            {fill (Y) circle (1pt);}
                            draw (A) -- (B) -- (C);
                            draw ($ (auxAB)!1.2cm!90:(B) $) -- ($ (auxAB)!1.2cm!-90:(B) $);
                            draw ($ (auxBC)!1.2cm!90:(B) $) -- ($ (auxBC)!1.2cm!-90:(B) $);
                            node[anchor=north,text width=7cm] (int) at (2.5,0){The center of the circle is
                            then simply determined by
                            [gamma_1(alpha)~=~gamma_2(beta);, ]
                            which has the solution
                            [
                            alpha~=~frac{-(x_2cdot x_3) + x_3cdot x_4 + y_3cdot (y_4-y_2 )}{x_3cdot y_1 - x_1cdot y_3};.
                            ]
                            This is texttt{textbackslash n1} in the Tiemph{k}Z style texttt{circle through 3 points}.
                            };
                            fi
                            ifnumX=5
                            draw[circle through 3 points={A}{B}{C}];
                            node[anchor=north,text width=7cm] (int) at (2.5,-0.1){Once we have the center,
                            determining the radius (texttt{textbackslash n2}) is trivial, and we can draw
                            the circle with a simple texttt{insert path}.};
                            fi
                            end{tikzpicture}}
                            end{document}


                            enter image description here







                            share|improve this answer














                            share|improve this answer



                            share|improve this answer








                            edited Nov 21 at 21:43

























                            answered Nov 21 at 19:07









                            marmot

                            78k487166




                            78k487166












                            • (+1) Give the analytical expression also, if possible :)
                              – nidhin
                              Nov 21 at 19:16










                            • @nidhin It is in the code, isn't it?
                              – marmot
                              Nov 21 at 19:20










                            • Yes it is there. I meant outside the code. Mathematical expression.
                              – nidhin
                              Nov 21 at 19:25










                            • @nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create.
                              – marmot
                              Nov 21 at 19:54










                            • Wow. That was more than expected. :)
                              – nidhin
                              Nov 21 at 20:03


















                            • (+1) Give the analytical expression also, if possible :)
                              – nidhin
                              Nov 21 at 19:16










                            • @nidhin It is in the code, isn't it?
                              – marmot
                              Nov 21 at 19:20










                            • Yes it is there. I meant outside the code. Mathematical expression.
                              – nidhin
                              Nov 21 at 19:25










                            • @nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create.
                              – marmot
                              Nov 21 at 19:54










                            • Wow. That was more than expected. :)
                              – nidhin
                              Nov 21 at 20:03
















                            (+1) Give the analytical expression also, if possible :)
                            – nidhin
                            Nov 21 at 19:16




                            (+1) Give the analytical expression also, if possible :)
                            – nidhin
                            Nov 21 at 19:16












                            @nidhin It is in the code, isn't it?
                            – marmot
                            Nov 21 at 19:20




                            @nidhin It is in the code, isn't it?
                            – marmot
                            Nov 21 at 19:20












                            Yes it is there. I meant outside the code. Mathematical expression.
                            – nidhin
                            Nov 21 at 19:25




                            Yes it is there. I meant outside the code. Mathematical expression.
                            – nidhin
                            Nov 21 at 19:25












                            @nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create.
                            – marmot
                            Nov 21 at 19:54




                            @nidhin Done. There is a simple animation that you can create with convert -delay 800 -loop 0 -density 300 -alpha remove <pdf> <gif>, where pdf is the pdf file that gets created if you compile the lower code, and gif the name of the animated gif you will create.
                            – marmot
                            Nov 21 at 19:54












                            Wow. That was more than expected. :)
                            – nidhin
                            Nov 21 at 20:03




                            Wow. That was more than expected. :)
                            – nidhin
                            Nov 21 at 20:03










                            up vote
                            7
                            down vote













                            You can use tkz-euclide like this:



                            documentclass{standalone}
                            usepackage{tikz}
                            usepackage{tkz-euclide}
                            usetikzlibrary{calc,through}
                            begin{document}
                            begin{tikzpicture}
                            coordinate (A) at (1,1);
                            coordinate (B) at (2,2);
                            coordinate (C) at (3,1.5);

                            node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                            foreach i in {A,B,C} {
                            node[circle,minimum size=1pt,fill=red] at(i) {};
                            }
                            tkzCircumCenter(A,B,C)tkzGetPoint{O}
                            tkzDrawCircle(O,A)
                            end{tikzpicture}
                            end{document}


                            (Modified from https://tex.stackexchange.com/a/16024/8650)
                            Circles



                            If you chose to use tkz-euclide, then you should consider to do all of your drawing with it - depending on what it is - you can e.g. define your points with tkzDefPoint(x,y).






                            share|improve this answer

























                              up vote
                              7
                              down vote













                              You can use tkz-euclide like this:



                              documentclass{standalone}
                              usepackage{tikz}
                              usepackage{tkz-euclide}
                              usetikzlibrary{calc,through}
                              begin{document}
                              begin{tikzpicture}
                              coordinate (A) at (1,1);
                              coordinate (B) at (2,2);
                              coordinate (C) at (3,1.5);

                              node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                              foreach i in {A,B,C} {
                              node[circle,minimum size=1pt,fill=red] at(i) {};
                              }
                              tkzCircumCenter(A,B,C)tkzGetPoint{O}
                              tkzDrawCircle(O,A)
                              end{tikzpicture}
                              end{document}


                              (Modified from https://tex.stackexchange.com/a/16024/8650)
                              Circles



                              If you chose to use tkz-euclide, then you should consider to do all of your drawing with it - depending on what it is - you can e.g. define your points with tkzDefPoint(x,y).






                              share|improve this answer























                                up vote
                                7
                                down vote










                                up vote
                                7
                                down vote









                                You can use tkz-euclide like this:



                                documentclass{standalone}
                                usepackage{tikz}
                                usepackage{tkz-euclide}
                                usetikzlibrary{calc,through}
                                begin{document}
                                begin{tikzpicture}
                                coordinate (A) at (1,1);
                                coordinate (B) at (2,2);
                                coordinate (C) at (3,1.5);

                                node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                                foreach i in {A,B,C} {
                                node[circle,minimum size=1pt,fill=red] at(i) {};
                                }
                                tkzCircumCenter(A,B,C)tkzGetPoint{O}
                                tkzDrawCircle(O,A)
                                end{tikzpicture}
                                end{document}


                                (Modified from https://tex.stackexchange.com/a/16024/8650)
                                Circles



                                If you chose to use tkz-euclide, then you should consider to do all of your drawing with it - depending on what it is - you can e.g. define your points with tkzDefPoint(x,y).






                                share|improve this answer












                                You can use tkz-euclide like this:



                                documentclass{standalone}
                                usepackage{tikz}
                                usepackage{tkz-euclide}
                                usetikzlibrary{calc,through}
                                begin{document}
                                begin{tikzpicture}
                                coordinate (A) at (1,1);
                                coordinate (B) at (2,2);
                                coordinate (C) at (3,1.5);

                                node[draw,line width=2pt] [circle through={(A)(B)(C)}] {};

                                foreach i in {A,B,C} {
                                node[circle,minimum size=1pt,fill=red] at(i) {};
                                }
                                tkzCircumCenter(A,B,C)tkzGetPoint{O}
                                tkzDrawCircle(O,A)
                                end{tikzpicture}
                                end{document}


                                (Modified from https://tex.stackexchange.com/a/16024/8650)
                                Circles



                                If you chose to use tkz-euclide, then you should consider to do all of your drawing with it - depending on what it is - you can e.g. define your points with tkzDefPoint(x,y).







                                share|improve this answer












                                share|improve this answer



                                share|improve this answer










                                answered Nov 21 at 17:52









                                hpekristiansen

                                4,87862863




                                4,87862863






















                                    up vote
                                    7
                                    down vote













                                    Just for comparison purpose.



                                    documentclass[pstricks]{standalone}  
                                    usepackage{pst-eucl}
                                    begin{document}
                                    foreach i in {1.0,1.2,...,4.0}{
                                    begin{pspicture}(-5,-5)(5,5)
                                    pstTriangle(4;30){A}(4;90){B}(i;-45){C}
                                    pstCircleABC{A}{B}{C}{O}
                                    end{pspicture}}
                                    end{document}


                                    enter image description here






                                    share|improve this answer

























                                      up vote
                                      7
                                      down vote













                                      Just for comparison purpose.



                                      documentclass[pstricks]{standalone}  
                                      usepackage{pst-eucl}
                                      begin{document}
                                      foreach i in {1.0,1.2,...,4.0}{
                                      begin{pspicture}(-5,-5)(5,5)
                                      pstTriangle(4;30){A}(4;90){B}(i;-45){C}
                                      pstCircleABC{A}{B}{C}{O}
                                      end{pspicture}}
                                      end{document}


                                      enter image description here






                                      share|improve this answer























                                        up vote
                                        7
                                        down vote










                                        up vote
                                        7
                                        down vote









                                        Just for comparison purpose.



                                        documentclass[pstricks]{standalone}  
                                        usepackage{pst-eucl}
                                        begin{document}
                                        foreach i in {1.0,1.2,...,4.0}{
                                        begin{pspicture}(-5,-5)(5,5)
                                        pstTriangle(4;30){A}(4;90){B}(i;-45){C}
                                        pstCircleABC{A}{B}{C}{O}
                                        end{pspicture}}
                                        end{document}


                                        enter image description here






                                        share|improve this answer












                                        Just for comparison purpose.



                                        documentclass[pstricks]{standalone}  
                                        usepackage{pst-eucl}
                                        begin{document}
                                        foreach i in {1.0,1.2,...,4.0}{
                                        begin{pspicture}(-5,-5)(5,5)
                                        pstTriangle(4;30){A}(4;90){B}(i;-45){C}
                                        pstCircleABC{A}{B}{C}{O}
                                        end{pspicture}}
                                        end{document}


                                        enter image description here







                                        share|improve this answer












                                        share|improve this answer



                                        share|improve this answer










                                        answered Nov 21 at 18:00









                                        Artificial Stupidity

                                        4,8191832




                                        4,8191832






















                                            up vote
                                            6
                                            down vote













                                            The code



                                            node [draw] at (1,1) [circle through={(A)}] {};


                                            draw a circle whose center is at (1,1) and passes through A. In this case the center of the Circumscribed circle has to be calculated before using through.



                                            I just used the Straightedge and compass construction to calculate the center and then drew circle. The basic idea is that all the perpendicular bisectors of the edges of a triangle meet at the same point: the circumcenter.



                                            enter image description here



                                            documentclass{standalone}
                                            usepackage{tikz}
                                            usetikzlibrary{through,intersections}
                                            begin{document}
                                            begin{tikzpicture}
                                            coordinate (A) at (1,1);
                                            coordinate (B) at (2,2);
                                            coordinate (C) at (3,1.5);
                                            path[name path=c1] (A) circle[radius=5cm];
                                            path[name path=c2] (B) circle[radius=5cm];
                                            path[name path=c3] (C) circle[radius=5cm];
                                            path[name intersections={of = c1 and c2}];
                                            path[name path=o1] (intersection-1)--(intersection-2);
                                            path[name intersections={of = c2 and c3}];
                                            path[name path=o2] (intersection-1)--(intersection-2);
                                            path[name intersections={of = o1 and o2}];
                                            node[draw,line width=2pt] at (intersection-1) [circle through={(A)}]{};

                                            foreach i in {A,B,C} {
                                            node[circle,minimum size=1pt,fill=red] at(i) {};
                                            }
                                            end{tikzpicture}
                                            end{document}





                                            share|improve this answer



























                                              up vote
                                              6
                                              down vote













                                              The code



                                              node [draw] at (1,1) [circle through={(A)}] {};


                                              draw a circle whose center is at (1,1) and passes through A. In this case the center of the Circumscribed circle has to be calculated before using through.



                                              I just used the Straightedge and compass construction to calculate the center and then drew circle. The basic idea is that all the perpendicular bisectors of the edges of a triangle meet at the same point: the circumcenter.



                                              enter image description here



                                              documentclass{standalone}
                                              usepackage{tikz}
                                              usetikzlibrary{through,intersections}
                                              begin{document}
                                              begin{tikzpicture}
                                              coordinate (A) at (1,1);
                                              coordinate (B) at (2,2);
                                              coordinate (C) at (3,1.5);
                                              path[name path=c1] (A) circle[radius=5cm];
                                              path[name path=c2] (B) circle[radius=5cm];
                                              path[name path=c3] (C) circle[radius=5cm];
                                              path[name intersections={of = c1 and c2}];
                                              path[name path=o1] (intersection-1)--(intersection-2);
                                              path[name intersections={of = c2 and c3}];
                                              path[name path=o2] (intersection-1)--(intersection-2);
                                              path[name intersections={of = o1 and o2}];
                                              node[draw,line width=2pt] at (intersection-1) [circle through={(A)}]{};

                                              foreach i in {A,B,C} {
                                              node[circle,minimum size=1pt,fill=red] at(i) {};
                                              }
                                              end{tikzpicture}
                                              end{document}





                                              share|improve this answer

























                                                up vote
                                                6
                                                down vote










                                                up vote
                                                6
                                                down vote









                                                The code



                                                node [draw] at (1,1) [circle through={(A)}] {};


                                                draw a circle whose center is at (1,1) and passes through A. In this case the center of the Circumscribed circle has to be calculated before using through.



                                                I just used the Straightedge and compass construction to calculate the center and then drew circle. The basic idea is that all the perpendicular bisectors of the edges of a triangle meet at the same point: the circumcenter.



                                                enter image description here



                                                documentclass{standalone}
                                                usepackage{tikz}
                                                usetikzlibrary{through,intersections}
                                                begin{document}
                                                begin{tikzpicture}
                                                coordinate (A) at (1,1);
                                                coordinate (B) at (2,2);
                                                coordinate (C) at (3,1.5);
                                                path[name path=c1] (A) circle[radius=5cm];
                                                path[name path=c2] (B) circle[radius=5cm];
                                                path[name path=c3] (C) circle[radius=5cm];
                                                path[name intersections={of = c1 and c2}];
                                                path[name path=o1] (intersection-1)--(intersection-2);
                                                path[name intersections={of = c2 and c3}];
                                                path[name path=o2] (intersection-1)--(intersection-2);
                                                path[name intersections={of = o1 and o2}];
                                                node[draw,line width=2pt] at (intersection-1) [circle through={(A)}]{};

                                                foreach i in {A,B,C} {
                                                node[circle,minimum size=1pt,fill=red] at(i) {};
                                                }
                                                end{tikzpicture}
                                                end{document}





                                                share|improve this answer














                                                The code



                                                node [draw] at (1,1) [circle through={(A)}] {};


                                                draw a circle whose center is at (1,1) and passes through A. In this case the center of the Circumscribed circle has to be calculated before using through.



                                                I just used the Straightedge and compass construction to calculate the center and then drew circle. The basic idea is that all the perpendicular bisectors of the edges of a triangle meet at the same point: the circumcenter.



                                                enter image description here



                                                documentclass{standalone}
                                                usepackage{tikz}
                                                usetikzlibrary{through,intersections}
                                                begin{document}
                                                begin{tikzpicture}
                                                coordinate (A) at (1,1);
                                                coordinate (B) at (2,2);
                                                coordinate (C) at (3,1.5);
                                                path[name path=c1] (A) circle[radius=5cm];
                                                path[name path=c2] (B) circle[radius=5cm];
                                                path[name path=c3] (C) circle[radius=5cm];
                                                path[name intersections={of = c1 and c2}];
                                                path[name path=o1] (intersection-1)--(intersection-2);
                                                path[name intersections={of = c2 and c3}];
                                                path[name path=o2] (intersection-1)--(intersection-2);
                                                path[name intersections={of = o1 and o2}];
                                                node[draw,line width=2pt] at (intersection-1) [circle through={(A)}]{};

                                                foreach i in {A,B,C} {
                                                node[circle,minimum size=1pt,fill=red] at(i) {};
                                                }
                                                end{tikzpicture}
                                                end{document}






                                                share|improve this answer














                                                share|improve this answer



                                                share|improve this answer








                                                edited Nov 21 at 19:05

























                                                answered Nov 21 at 18:32









                                                nidhin

                                                1,800921




                                                1,800921






















                                                    up vote
                                                    5
                                                    down vote













                                                    Just for fun, another solution inspired by the @marmot solution that calculates the intersection of two defined perpendicular bisector with the calc library.



                                                    cercle



                                                    documentclass[tikz,border=5mm]{standalone}
                                                    usetikzlibrary{calc,through}
                                                    tikzset{circle through 3 points/.style n args={3}{%
                                                    insert path={let p1=($(#1)!0.5!(#2)$),
                                                    p2=($(#1)!0.5!(#3)$),
                                                    p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                                                    p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                                                    p5=(intersection of p1--p3 and p2--p4)
                                                    in
                                                    node at (p5) [draw,line width=2pt,circle through= {(#1)}]{}}
                                                    }}

                                                    begin{document}
                                                    begin{tikzpicture}
                                                    coordinate (A) at (1,1);
                                                    coordinate (B) at (2,2);
                                                    coordinate (C) at (3,1.5);
                                                    draw[circle through 3 points={A}{B}{C}];
                                                    foreach i in {A,B,C} {
                                                    node[circle,minimum size=1pt,fill=red] at(i) {};
                                                    }
                                                    end{tikzpicture}
                                                    end{document}





                                                    share|improve this answer





















                                                    • Nice! I didn't know that one could use p1 etc. in intersection of p1--p3 and p2--p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what path usually does.
                                                      – marmot
                                                      Nov 21 at 21:11










                                                    • @marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
                                                      – AndréC
                                                      Nov 21 at 21:56












                                                    • I really could only do that after I learned the intersection of p1--p3 and p2--p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
                                                      – marmot
                                                      Nov 21 at 22:03










                                                    • @marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
                                                      – AndréC
                                                      Nov 21 at 22:07










                                                    • I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
                                                      – marmot
                                                      Nov 21 at 22:11















                                                    up vote
                                                    5
                                                    down vote













                                                    Just for fun, another solution inspired by the @marmot solution that calculates the intersection of two defined perpendicular bisector with the calc library.



                                                    cercle



                                                    documentclass[tikz,border=5mm]{standalone}
                                                    usetikzlibrary{calc,through}
                                                    tikzset{circle through 3 points/.style n args={3}{%
                                                    insert path={let p1=($(#1)!0.5!(#2)$),
                                                    p2=($(#1)!0.5!(#3)$),
                                                    p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                                                    p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                                                    p5=(intersection of p1--p3 and p2--p4)
                                                    in
                                                    node at (p5) [draw,line width=2pt,circle through= {(#1)}]{}}
                                                    }}

                                                    begin{document}
                                                    begin{tikzpicture}
                                                    coordinate (A) at (1,1);
                                                    coordinate (B) at (2,2);
                                                    coordinate (C) at (3,1.5);
                                                    draw[circle through 3 points={A}{B}{C}];
                                                    foreach i in {A,B,C} {
                                                    node[circle,minimum size=1pt,fill=red] at(i) {};
                                                    }
                                                    end{tikzpicture}
                                                    end{document}





                                                    share|improve this answer





















                                                    • Nice! I didn't know that one could use p1 etc. in intersection of p1--p3 and p2--p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what path usually does.
                                                      – marmot
                                                      Nov 21 at 21:11










                                                    • @marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
                                                      – AndréC
                                                      Nov 21 at 21:56












                                                    • I really could only do that after I learned the intersection of p1--p3 and p2--p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
                                                      – marmot
                                                      Nov 21 at 22:03










                                                    • @marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
                                                      – AndréC
                                                      Nov 21 at 22:07










                                                    • I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
                                                      – marmot
                                                      Nov 21 at 22:11













                                                    up vote
                                                    5
                                                    down vote










                                                    up vote
                                                    5
                                                    down vote









                                                    Just for fun, another solution inspired by the @marmot solution that calculates the intersection of two defined perpendicular bisector with the calc library.



                                                    cercle



                                                    documentclass[tikz,border=5mm]{standalone}
                                                    usetikzlibrary{calc,through}
                                                    tikzset{circle through 3 points/.style n args={3}{%
                                                    insert path={let p1=($(#1)!0.5!(#2)$),
                                                    p2=($(#1)!0.5!(#3)$),
                                                    p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                                                    p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                                                    p5=(intersection of p1--p3 and p2--p4)
                                                    in
                                                    node at (p5) [draw,line width=2pt,circle through= {(#1)}]{}}
                                                    }}

                                                    begin{document}
                                                    begin{tikzpicture}
                                                    coordinate (A) at (1,1);
                                                    coordinate (B) at (2,2);
                                                    coordinate (C) at (3,1.5);
                                                    draw[circle through 3 points={A}{B}{C}];
                                                    foreach i in {A,B,C} {
                                                    node[circle,minimum size=1pt,fill=red] at(i) {};
                                                    }
                                                    end{tikzpicture}
                                                    end{document}





                                                    share|improve this answer












                                                    Just for fun, another solution inspired by the @marmot solution that calculates the intersection of two defined perpendicular bisector with the calc library.



                                                    cercle



                                                    documentclass[tikz,border=5mm]{standalone}
                                                    usetikzlibrary{calc,through}
                                                    tikzset{circle through 3 points/.style n args={3}{%
                                                    insert path={let p1=($(#1)!0.5!(#2)$),
                                                    p2=($(#1)!0.5!(#3)$),
                                                    p3=($(#1)!0.5!(#2)!1!-90:(#2)$),
                                                    p4=($(#1)!0.5!(#3)!1!90:(#3)$),
                                                    p5=(intersection of p1--p3 and p2--p4)
                                                    in
                                                    node at (p5) [draw,line width=2pt,circle through= {(#1)}]{}}
                                                    }}

                                                    begin{document}
                                                    begin{tikzpicture}
                                                    coordinate (A) at (1,1);
                                                    coordinate (B) at (2,2);
                                                    coordinate (C) at (3,1.5);
                                                    draw[circle through 3 points={A}{B}{C}];
                                                    foreach i in {A,B,C} {
                                                    node[circle,minimum size=1pt,fill=red] at(i) {};
                                                    }
                                                    end{tikzpicture}
                                                    end{document}






                                                    share|improve this answer












                                                    share|improve this answer



                                                    share|improve this answer










                                                    answered Nov 21 at 20:57









                                                    AndréC

                                                    6,24711140




                                                    6,24711140












                                                    • Nice! I didn't know that one could use p1 etc. in intersection of p1--p3 and p2--p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what path usually does.
                                                      – marmot
                                                      Nov 21 at 21:11










                                                    • @marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
                                                      – AndréC
                                                      Nov 21 at 21:56












                                                    • I really could only do that after I learned the intersection of p1--p3 and p2--p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
                                                      – marmot
                                                      Nov 21 at 22:03










                                                    • @marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
                                                      – AndréC
                                                      Nov 21 at 22:07










                                                    • I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
                                                      – marmot
                                                      Nov 21 at 22:11


















                                                    • Nice! I didn't know that one could use p1 etc. in intersection of p1--p3 and p2--p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what path usually does.
                                                      – marmot
                                                      Nov 21 at 21:11










                                                    • @marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
                                                      – AndréC
                                                      Nov 21 at 21:56












                                                    • I really could only do that after I learned the intersection of p1--p3 and p2--p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
                                                      – marmot
                                                      Nov 21 at 22:03










                                                    • @marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
                                                      – AndréC
                                                      Nov 21 at 22:07










                                                    • I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
                                                      – marmot
                                                      Nov 21 at 22:11
















                                                    Nice! I didn't know that one could use p1 etc. in intersection of p1--p3 and p2--p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what path usually does.
                                                    – marmot
                                                    Nov 21 at 21:11




                                                    Nice! I didn't know that one could use p1 etc. in intersection of p1--p3 and p2--p4. (I guess it will be more elegant if you manage to eliminate draw,line width=2pt, from the node since path[circle through 3 points={A}{B}{C}]; also draws the path, which is not what path usually does.
                                                    – marmot
                                                    Nov 21 at 21:11












                                                    @marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
                                                    – AndréC
                                                    Nov 21 at 21:56






                                                    @marmot It is simple to delete line width=2pt which is a given parameter of the lucky1928 question. On the other hand, it is more difficult to delete draw. To do that, you have to find a way to pass the parameters to the node, I'll get down to work and explore (finally!) the key handlers. I just saw that you succeeded! Congratulations!
                                                    – AndréC
                                                    Nov 21 at 21:56














                                                    I really could only do that after I learned the intersection of p1--p3 and p2--p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
                                                    – marmot
                                                    Nov 21 at 22:03




                                                    I really could only do that after I learned the intersection of p1--p3 and p2--p4 thing from you, which is much better than the analytic computation of the intersection point, and presumably also more stable (even though the analytic derivation was fun ;-). Thanks a lot! (My aim was to create a style that behaves the same way as circle through, which your observation made possible, so thanks again!)
                                                    – marmot
                                                    Nov 21 at 22:03












                                                    @marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
                                                    – AndréC
                                                    Nov 21 at 22:07




                                                    @marmot Thank you very much and I have to return the compliment because it is by adapting your solution that I learned how to use insert path and that I am finally motivated to learn key handlers.
                                                    – AndréC
                                                    Nov 21 at 22:07












                                                    I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
                                                    – marmot
                                                    Nov 21 at 22:11




                                                    I really like that you acknowledge other posts (and I try to do the same). I wish all users would do that.... ;-)
                                                    – marmot
                                                    Nov 21 at 22:11










                                                    up vote
                                                    5
                                                    down vote













                                                    Just for fun with @AndréC's answer:



                                                    documentclass[tikz,border=5mm]{standalone}
                                                    usetikzlibrary{calc,through}
                                                    begin{document}
                                                    begin{tikzpicture}
                                                    coordinate (A) at (1,1);
                                                    coordinate (B) at (2,2);
                                                    coordinate (C) at (3,1.5);

                                                    draw let p1=($(A)!0.5!(B)$),
                                                    p2=($(A)!0.5!(C)$),
                                                    p3=($(p1)!2!-90:(B)$),
                                                    p4=($(p1)!2!90:(B)$),
                                                    p5=($(p2)!2!-90:(C)$),
                                                    p6=($(p2)!2!90:(C)$),
                                                    p7=(intersection of p3--p4 and p5--p6)
                                                    in
                                                    (A) -- (B)
                                                    (A) -- (C)
                                                    (p3) -- (p4)
                                                    (p5) -- (p6)
                                                    foreach j in {1,...,7} {
                                                    node[circle,minimum size=2pt,fill=red,inner sep=0,label=j] at(pj) {}
                                                    }
                                                    node[draw,line width=1pt,circle through= {(A)}] at (p7) {};

                                                    foreach i in {A,B,C} {
                                                    node[circle,minimum size=5pt,fill=red,inner sep=0,label=i] at(i) {};
                                                    }
                                                    end{tikzpicture}
                                                    end{document}


                                                    enter image description here






                                                    share|improve this answer



























                                                      up vote
                                                      5
                                                      down vote













                                                      Just for fun with @AndréC's answer:



                                                      documentclass[tikz,border=5mm]{standalone}
                                                      usetikzlibrary{calc,through}
                                                      begin{document}
                                                      begin{tikzpicture}
                                                      coordinate (A) at (1,1);
                                                      coordinate (B) at (2,2);
                                                      coordinate (C) at (3,1.5);

                                                      draw let p1=($(A)!0.5!(B)$),
                                                      p2=($(A)!0.5!(C)$),
                                                      p3=($(p1)!2!-90:(B)$),
                                                      p4=($(p1)!2!90:(B)$),
                                                      p5=($(p2)!2!-90:(C)$),
                                                      p6=($(p2)!2!90:(C)$),
                                                      p7=(intersection of p3--p4 and p5--p6)
                                                      in
                                                      (A) -- (B)
                                                      (A) -- (C)
                                                      (p3) -- (p4)
                                                      (p5) -- (p6)
                                                      foreach j in {1,...,7} {
                                                      node[circle,minimum size=2pt,fill=red,inner sep=0,label=j] at(pj) {}
                                                      }
                                                      node[draw,line width=1pt,circle through= {(A)}] at (p7) {};

                                                      foreach i in {A,B,C} {
                                                      node[circle,minimum size=5pt,fill=red,inner sep=0,label=i] at(i) {};
                                                      }
                                                      end{tikzpicture}
                                                      end{document}


                                                      enter image description here






                                                      share|improve this answer

























                                                        up vote
                                                        5
                                                        down vote










                                                        up vote
                                                        5
                                                        down vote









                                                        Just for fun with @AndréC's answer:



                                                        documentclass[tikz,border=5mm]{standalone}
                                                        usetikzlibrary{calc,through}
                                                        begin{document}
                                                        begin{tikzpicture}
                                                        coordinate (A) at (1,1);
                                                        coordinate (B) at (2,2);
                                                        coordinate (C) at (3,1.5);

                                                        draw let p1=($(A)!0.5!(B)$),
                                                        p2=($(A)!0.5!(C)$),
                                                        p3=($(p1)!2!-90:(B)$),
                                                        p4=($(p1)!2!90:(B)$),
                                                        p5=($(p2)!2!-90:(C)$),
                                                        p6=($(p2)!2!90:(C)$),
                                                        p7=(intersection of p3--p4 and p5--p6)
                                                        in
                                                        (A) -- (B)
                                                        (A) -- (C)
                                                        (p3) -- (p4)
                                                        (p5) -- (p6)
                                                        foreach j in {1,...,7} {
                                                        node[circle,minimum size=2pt,fill=red,inner sep=0,label=j] at(pj) {}
                                                        }
                                                        node[draw,line width=1pt,circle through= {(A)}] at (p7) {};

                                                        foreach i in {A,B,C} {
                                                        node[circle,minimum size=5pt,fill=red,inner sep=0,label=i] at(i) {};
                                                        }
                                                        end{tikzpicture}
                                                        end{document}


                                                        enter image description here






                                                        share|improve this answer














                                                        Just for fun with @AndréC's answer:



                                                        documentclass[tikz,border=5mm]{standalone}
                                                        usetikzlibrary{calc,through}
                                                        begin{document}
                                                        begin{tikzpicture}
                                                        coordinate (A) at (1,1);
                                                        coordinate (B) at (2,2);
                                                        coordinate (C) at (3,1.5);

                                                        draw let p1=($(A)!0.5!(B)$),
                                                        p2=($(A)!0.5!(C)$),
                                                        p3=($(p1)!2!-90:(B)$),
                                                        p4=($(p1)!2!90:(B)$),
                                                        p5=($(p2)!2!-90:(C)$),
                                                        p6=($(p2)!2!90:(C)$),
                                                        p7=(intersection of p3--p4 and p5--p6)
                                                        in
                                                        (A) -- (B)
                                                        (A) -- (C)
                                                        (p3) -- (p4)
                                                        (p5) -- (p6)
                                                        foreach j in {1,...,7} {
                                                        node[circle,minimum size=2pt,fill=red,inner sep=0,label=j] at(pj) {}
                                                        }
                                                        node[draw,line width=1pt,circle through= {(A)}] at (p7) {};

                                                        foreach i in {A,B,C} {
                                                        node[circle,minimum size=5pt,fill=red,inner sep=0,label=i] at(i) {};
                                                        }
                                                        end{tikzpicture}
                                                        end{document}


                                                        enter image description here







                                                        share|improve this answer














                                                        share|improve this answer



                                                        share|improve this answer








                                                        edited Nov 21 at 22:26

























                                                        answered Nov 21 at 22:14









                                                        beetlej

                                                        45429




                                                        45429






























                                                             

                                                            draft saved


                                                            draft discarded



















































                                                             


                                                            draft saved


                                                            draft discarded














                                                            StackExchange.ready(
                                                            function () {
                                                            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2ftex.stackexchange.com%2fquestions%2f461161%2fdrawing-a-circle-through-3-non-collinear-points%23new-answer', 'question_page');
                                                            }
                                                            );

                                                            Post as a guest















                                                            Required, but never shown





















































                                                            Required, but never shown














                                                            Required, but never shown












                                                            Required, but never shown







                                                            Required, but never shown

































                                                            Required, but never shown














                                                            Required, but never shown












                                                            Required, but never shown







                                                            Required, but never shown







                                                            Popular posts from this blog

                                                            AnyDesk - Fatal Program Failure

                                                            How to calibrate 16:9 built-in touch-screen to a 4:3 resolution?

                                                            QoS: MAC-Priority for clients behind a repeater