Tangent points of a circle touching two functions











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I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!



See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.



If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.



enter image description here



Thank you!



BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.










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    up vote
    1
    down vote

    favorite












    I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!



    See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.



    If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.



    enter image description here



    Thank you!



    BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.










    share|cite|improve this question









    New contributor




    Sejin Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
    Check out our Code of Conduct.






















      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!



      See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.



      If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.



      enter image description here



      Thank you!



      BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.










      share|cite|improve this question









      New contributor




      Sejin Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.











      I'm really confused with this and don't quite know where to start, and was hoping that I could get a booster in the right direction!



      See the picture, but here's a written description: A graph has two graphs plotted on it, $x=ln(x)$ and $x=ln(-x)$, to make a funnel shape. A circle is resting inside of that funnel, with a radius of $2$. How can I find a) the center point of the circle (which rests on the $y$-axis, I just need to know the $y$- value), and b) the tangent points or points of contact between the circle and the two plotted functions.



      If you're looking at the graph, then it would be points $P$ and $Q$ for the tangent points and point $R$ for the center of the circle.



      enter image description here



      Thank you!



      BTW: please don't just give the answer! I just need a pointer on what to find and where to go from there.







      calculus






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      New contributor




      Sejin Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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      share|cite|improve this question








      edited 18 hours ago









      Robert Z

      89.3k1056128




      89.3k1056128






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      asked 19 hours ago









      Sejin Kim

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      61




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      New contributor





      Sejin Kim is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
      Check out our Code of Conduct.






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      Check out our Code of Conduct.






















          3 Answers
          3






          active

          oldest

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          up vote
          0
          down vote













          HINT:
          I think the equations should be



          $$y=ln(x)........(1)$$
          $$y=ln(-x)........(2)$$
          $$(x-h)^2+(y-k)^2=4........(3)$$
          Now at tangent point the slope of the circle and the given graph will be the same.so,
          $$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$



          You will get a system of equations here..4 equations,4 unknown...
          Just solve them.






          share|cite|improve this answer




























            up vote
            0
            down vote













            HINTS:



            Draw normal/tangent at $Q$ cutting symmetry line at $R$



            Coordinates $$ Q:quad (x_1, log ,x_1) $$



            Slope is the derivative evaluated at that point.



            $$ tan phi= 1/x =1/x_1 $$



            Difference along central symmetry axis, vertical component of $QR$



            $$ h = x_1 cot phi $$



            Coordinates $$ R: (0,, log x_1 +h) $$



            Equation of circle tangent at $ (P,Q) $ centered at $R:$



            $$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$






            share|cite|improve this answer






























              up vote
              0
              down vote













              Start from the end and go back.
              Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
              The final step is to find $x_0>0$ such that $|RQ|=2$.



              Can you take it from here?






              share|cite|improve this answer























              • Is there a way to find the generic x-value algebraically, rather than graphically?
                – Sejin Kim
                18 mins ago











              Your Answer





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              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes








              up vote
              0
              down vote













              HINT:
              I think the equations should be



              $$y=ln(x)........(1)$$
              $$y=ln(-x)........(2)$$
              $$(x-h)^2+(y-k)^2=4........(3)$$
              Now at tangent point the slope of the circle and the given graph will be the same.so,
              $$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$



              You will get a system of equations here..4 equations,4 unknown...
              Just solve them.






              share|cite|improve this answer

























                up vote
                0
                down vote













                HINT:
                I think the equations should be



                $$y=ln(x)........(1)$$
                $$y=ln(-x)........(2)$$
                $$(x-h)^2+(y-k)^2=4........(3)$$
                Now at tangent point the slope of the circle and the given graph will be the same.so,
                $$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$



                You will get a system of equations here..4 equations,4 unknown...
                Just solve them.






                share|cite|improve this answer























                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  HINT:
                  I think the equations should be



                  $$y=ln(x)........(1)$$
                  $$y=ln(-x)........(2)$$
                  $$(x-h)^2+(y-k)^2=4........(3)$$
                  Now at tangent point the slope of the circle and the given graph will be the same.so,
                  $$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$



                  You will get a system of equations here..4 equations,4 unknown...
                  Just solve them.






                  share|cite|improve this answer












                  HINT:
                  I think the equations should be



                  $$y=ln(x)........(1)$$
                  $$y=ln(-x)........(2)$$
                  $$(x-h)^2+(y-k)^2=4........(3)$$
                  Now at tangent point the slope of the circle and the given graph will be the same.so,
                  $$dfrac{d}{dx}bigg((x-h)^2+(y-k)^2-4bigg)=dfrac{d}{dx}bigg(ln(pm x)bigg)........(4)$$



                  You will get a system of equations here..4 equations,4 unknown...
                  Just solve them.







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered 18 hours ago









                  Rakibul Islam Prince

                  80529




                  80529






















                      up vote
                      0
                      down vote













                      HINTS:



                      Draw normal/tangent at $Q$ cutting symmetry line at $R$



                      Coordinates $$ Q:quad (x_1, log ,x_1) $$



                      Slope is the derivative evaluated at that point.



                      $$ tan phi= 1/x =1/x_1 $$



                      Difference along central symmetry axis, vertical component of $QR$



                      $$ h = x_1 cot phi $$



                      Coordinates $$ R: (0,, log x_1 +h) $$



                      Equation of circle tangent at $ (P,Q) $ centered at $R:$



                      $$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$






                      share|cite|improve this answer



























                        up vote
                        0
                        down vote













                        HINTS:



                        Draw normal/tangent at $Q$ cutting symmetry line at $R$



                        Coordinates $$ Q:quad (x_1, log ,x_1) $$



                        Slope is the derivative evaluated at that point.



                        $$ tan phi= 1/x =1/x_1 $$



                        Difference along central symmetry axis, vertical component of $QR$



                        $$ h = x_1 cot phi $$



                        Coordinates $$ R: (0,, log x_1 +h) $$



                        Equation of circle tangent at $ (P,Q) $ centered at $R:$



                        $$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$






                        share|cite|improve this answer

























                          up vote
                          0
                          down vote










                          up vote
                          0
                          down vote









                          HINTS:



                          Draw normal/tangent at $Q$ cutting symmetry line at $R$



                          Coordinates $$ Q:quad (x_1, log ,x_1) $$



                          Slope is the derivative evaluated at that point.



                          $$ tan phi= 1/x =1/x_1 $$



                          Difference along central symmetry axis, vertical component of $QR$



                          $$ h = x_1 cot phi $$



                          Coordinates $$ R: (0,, log x_1 +h) $$



                          Equation of circle tangent at $ (P,Q) $ centered at $R:$



                          $$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$






                          share|cite|improve this answer














                          HINTS:



                          Draw normal/tangent at $Q$ cutting symmetry line at $R$



                          Coordinates $$ Q:quad (x_1, log ,x_1) $$



                          Slope is the derivative evaluated at that point.



                          $$ tan phi= 1/x =1/x_1 $$



                          Difference along central symmetry axis, vertical component of $QR$



                          $$ h = x_1 cot phi $$



                          Coordinates $$ R: (0,, log x_1 +h) $$



                          Equation of circle tangent at $ (P,Q) $ centered at $R:$



                          $$ (x- 0)^2 + [y-(log, x_1 +h)]^2 = (h/sin phi)^2 $$







                          share|cite|improve this answer














                          share|cite|improve this answer



                          share|cite|improve this answer








                          edited 17 hours ago

























                          answered 18 hours ago









                          Narasimham

                          20.4k52158




                          20.4k52158






















                              up vote
                              0
                              down vote













                              Start from the end and go back.
                              Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
                              The final step is to find $x_0>0$ such that $|RQ|=2$.



                              Can you take it from here?






                              share|cite|improve this answer























                              • Is there a way to find the generic x-value algebraically, rather than graphically?
                                – Sejin Kim
                                18 mins ago















                              up vote
                              0
                              down vote













                              Start from the end and go back.
                              Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
                              The final step is to find $x_0>0$ such that $|RQ|=2$.



                              Can you take it from here?






                              share|cite|improve this answer























                              • Is there a way to find the generic x-value algebraically, rather than graphically?
                                – Sejin Kim
                                18 mins ago













                              up vote
                              0
                              down vote










                              up vote
                              0
                              down vote









                              Start from the end and go back.
                              Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
                              The final step is to find $x_0>0$ such that $|RQ|=2$.



                              Can you take it from here?






                              share|cite|improve this answer














                              Start from the end and go back.
                              Note that the segment $RQ$ is orthogonal to the tangent line to the graph of $ln(x)$ at the point $Q$. So given a generic $x_0>0$, find the tangent line at $Q=(x_0,ln(x_0))$, the orthogonal line and determine $R$ which is along the line $x=0$.
                              The final step is to find $x_0>0$ such that $|RQ|=2$.



                              Can you take it from here?







                              share|cite|improve this answer














                              share|cite|improve this answer



                              share|cite|improve this answer








                              edited 17 hours ago

























                              answered 19 hours ago









                              Robert Z

                              89.3k1056128




                              89.3k1056128












                              • Is there a way to find the generic x-value algebraically, rather than graphically?
                                – Sejin Kim
                                18 mins ago


















                              • Is there a way to find the generic x-value algebraically, rather than graphically?
                                – Sejin Kim
                                18 mins ago
















                              Is there a way to find the generic x-value algebraically, rather than graphically?
                              – Sejin Kim
                              18 mins ago




                              Is there a way to find the generic x-value algebraically, rather than graphically?
                              – Sejin Kim
                              18 mins ago










                              Sejin Kim is a new contributor. Be nice, and check out our Code of Conduct.










                               

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                              Sejin Kim is a new contributor. Be nice, and check out our Code of Conduct.















                               


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