Find the equation of the circle given the tangent line, point on the circle, and the radius [closed]
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The problem was this: radius is $2$, tangent to the $x$ axis, passes through $(1, -1)$. I don't know how to solve this, and my math teacher didnt teach this yet.
circle
edited Nov 18 at 10:47
Yadati Kiran
1,243417
asked Nov 18 at 10:16
aki
41
closed as off-topic by amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos Nov 18 at 23:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
add a comment |
up vote
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down vote
favorite
The problem was this: radius is $2$, tangent to the $x$ axis, passes through $(1, -1)$. I don't know how to solve this, and my math teacher didnt teach this yet.
circle
edited Nov 18 at 10:47
Yadati Kiran
1,243417
asked Nov 18 at 10:16
aki
41
closed as off-topic by amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos Nov 18 at 23:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
"my math teacher didnt teach this yet" A teacher shouldn't have to teach the exact method of solving a specific problem in order to expect his / her students to solve it. You have likely been taught all the tools you need: You know what the equation of a circle looks like, and you've been taught how to solve equations. In this problem you've been given information you can enter directly into that equation in order to solve and figure out the parts you haven't been told. Try it out, see how far you get, then let us know where you're stuck.
– Arthur
Nov 18 at 10:25
The circle also passes through $(x,0)$.
– Yadati Kiran
Nov 18 at 10:49
$(x-h)^2 + (y-k)^2=2^2 $ How to find $(h,k)?$
– Narasimham
Nov 18 at 18:17
Hint: the center has coordinates $(a,2)$ or $(a,-2)$.
– egreg
Nov 18 at 23:30
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
The problem was this: radius is $2$, tangent to the $x$ axis, passes through $(1, -1)$. I don't know how to solve this, and my math teacher didnt teach this yet.
circle
edited Nov 18 at 10:47
Yadati Kiran
1,243417
asked Nov 18 at 10:16
aki
41
The problem was this: radius is $2$, tangent to the $x$ axis, passes through $(1, -1)$. I don't know how to solve this, and my math teacher didnt teach this yet.
circle
circle
edited Nov 18 at 10:47
Yadati Kiran
1,243417
asked Nov 18 at 10:16
aki
41
edited Nov 18 at 10:47
Yadati Kiran
1,243417
asked Nov 18 at 10:16
aki
41
edited Nov 18 at 10:47
Yadati Kiran
1,243417
edited Nov 18 at 10:47
Yadati Kiran
1,243417
edited Nov 18 at 10:47
Yadati Kiran
1,243417
1,243417
asked Nov 18 at 10:16
aki
41
asked Nov 18 at 10:16
aki
41
asked Nov 18 at 10:16
aki
41
41
closed as off-topic by amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos Nov 18 at 23:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
closed as off-topic by amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos Nov 18 at 23:29
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – amWhy, Brahadeesh, Paul Frost, T. Bongers, José Carlos Santos
If this question can be reworded to fit the rules in the help center, please edit the question.
"my math teacher didnt teach this yet" A teacher shouldn't have to teach the exact method of solving a specific problem in order to expect his / her students to solve it. You have likely been taught all the tools you need: You know what the equation of a circle looks like, and you've been taught how to solve equations. In this problem you've been given information you can enter directly into that equation in order to solve and figure out the parts you haven't been told. Try it out, see how far you get, then let us know where you're stuck.
– Arthur
Nov 18 at 10:25
The circle also passes through $(x,0)$.
– Yadati Kiran
Nov 18 at 10:49
$(x-h)^2 + (y-k)^2=2^2 $ How to find $(h,k)?$
– Narasimham
Nov 18 at 18:17
Hint: the center has coordinates $(a,2)$ or $(a,-2)$.
– egreg
Nov 18 at 23:30
add a comment |
"my math teacher didnt teach this yet" A teacher shouldn't have to teach the exact method of solving a specific problem in order to expect his / her students to solve it. You have likely been taught all the tools you need: You know what the equation of a circle looks like, and you've been taught how to solve equations. In this problem you've been given information you can enter directly into that equation in order to solve and figure out the parts you haven't been told. Try it out, see how far you get, then let us know where you're stuck.
– Arthur
Nov 18 at 10:25
The circle also passes through $(x,0)$.
– Yadati Kiran
Nov 18 at 10:49
$(x-h)^2 + (y-k)^2=2^2 $ How to find $(h,k)?$
– Narasimham
Nov 18 at 18:17
Hint: the center has coordinates $(a,2)$ or $(a,-2)$.
– egreg
Nov 18 at 23:30
"my math teacher didnt teach this yet" A teacher shouldn't have to teach the exact method of solving a specific problem in order to expect his / her students to solve it. You have likely been taught all the tools you need: You know what the equation of a circle looks like, and you've been taught how to solve equations. In this problem you've been given information you can enter directly into that equation in order to solve and figure out the parts you haven't been told. Try it out, see how far you get, then let us know where you're stuck.
– Arthur
Nov 18 at 10:25
"my math teacher didnt teach this yet" A teacher shouldn't have to teach the exact method of solving a specific problem in order to expect his / her students to solve it. You have likely been taught all the tools you need: You know what the equation of a circle looks like, and you've been taught how to solve equations. In this problem you've been given information you can enter directly into that equation in order to solve and figure out the parts you haven't been told. Try it out, see how far you get, then let us know where you're stuck.
– Arthur
Nov 18 at 10:25
The circle also passes through $(x,0)$.
– Yadati Kiran
Nov 18 at 10:49
The circle also passes through $(x,0)$.
– Yadati Kiran
Nov 18 at 10:49
$(x-h)^2 + (y-k)^2=2^2 $ How to find $(h,k)?$
– Narasimham
Nov 18 at 18:17
$(x-h)^2 + (y-k)^2=2^2 $ How to find $(h,k)?$
– Narasimham
Nov 18 at 18:17
Hint: the center has coordinates $(a,2)$ or $(a,-2)$.
– egreg
Nov 18 at 23:30
Hint: the center has coordinates $(a,2)$ or $(a,-2)$.
– egreg
Nov 18 at 23:30
add a comment |
2 Answers
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Given the conditions you gave, there is only two circle that works: their center is $left(begin{array}{cc} 1-sqrt{3}\ -2 end{array}right)$ and $left(begin{array}{cc} 1+sqrt{3}\ -2 end{array}right)$, their radius is $2$.
So their respective equation are:
$(x - 1+sqrt{3})^2 + (y+2)^2 = 4$ and $(x-1-sqrt{3})^2 + (y+2)^2 = 4$.
answered Nov 18 at 10:49
Euler Pythagoras
4249
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Since the circle is tangent to the x-axis and passes through $(1,-1)$ with radius $2$ we know the center is below the x-axis and $2$ units away from it.
Thus the center is a point $ C(x,-2)$ which is $2$ unit apart from $(1,-1)$
That gives us $$(x-1)^2 + (-1)^2 =4$$ which implies $x=1pm sqrt 3 $
There are two circles with equations $$(x-1pm sqrt 3)^2 +(y+2)^2=4$$
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
add a comment |
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
Given the conditions you gave, there is only two circle that works: their center is $left(begin{array}{cc} 1-sqrt{3}\ -2 end{array}right)$ and $left(begin{array}{cc} 1+sqrt{3}\ -2 end{array}right)$, their radius is $2$.
So their respective equation are:
$(x - 1+sqrt{3})^2 + (y+2)^2 = 4$ and $(x-1-sqrt{3})^2 + (y+2)^2 = 4$.
answered Nov 18 at 10:49
Euler Pythagoras
4249
add a comment |
up vote
1
down vote
Given the conditions you gave, there is only two circle that works: their center is $left(begin{array}{cc} 1-sqrt{3}\ -2 end{array}right)$ and $left(begin{array}{cc} 1+sqrt{3}\ -2 end{array}right)$, their radius is $2$.
So their respective equation are:
$(x - 1+sqrt{3})^2 + (y+2)^2 = 4$ and $(x-1-sqrt{3})^2 + (y+2)^2 = 4$.
answered Nov 18 at 10:49
Euler Pythagoras
4249
add a comment |
up vote
1
down vote
up vote
1
down vote
Given the conditions you gave, there is only two circle that works: their center is $left(begin{array}{cc} 1-sqrt{3}\ -2 end{array}right)$ and $left(begin{array}{cc} 1+sqrt{3}\ -2 end{array}right)$, their radius is $2$.
So their respective equation are:
$(x - 1+sqrt{3})^2 + (y+2)^2 = 4$ and $(x-1-sqrt{3})^2 + (y+2)^2 = 4$.
answered Nov 18 at 10:49
Euler Pythagoras
4249
Given the conditions you gave, there is only two circle that works: their center is $left(begin{array}{cc} 1-sqrt{3}\ -2 end{array}right)$ and $left(begin{array}{cc} 1+sqrt{3}\ -2 end{array}right)$, their radius is $2$.
So their respective equation are:
$(x - 1+sqrt{3})^2 + (y+2)^2 = 4$ and $(x-1-sqrt{3})^2 + (y+2)^2 = 4$.
answered Nov 18 at 10:49
Euler Pythagoras
4249
edited Nov 18 at 11:33
answered Nov 18 at 10:49
Euler Pythagoras
4249
answered Nov 18 at 10:49
Euler Pythagoras
4249
answered Nov 18 at 10:49
Euler Pythagoras
4249
4249
add a comment |
add a comment |
up vote
0
down vote
Since the circle is tangent to the x-axis and passes through $(1,-1)$ with radius $2$ we know the center is below the x-axis and $2$ units away from it.
Thus the center is a point $ C(x,-2)$ which is $2$ unit apart from $(1,-1)$
That gives us $$(x-1)^2 + (-1)^2 =4$$ which implies $x=1pm sqrt 3 $
There are two circles with equations $$(x-1pm sqrt 3)^2 +(y+2)^2=4$$
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
Since the circle is tangent to the x-axis and passes through $(1,-1)$ with radius $2$ we know the center is below the x-axis and $2$ units away from it.
Thus the center is a point $ C(x,-2)$ which is $2$ unit apart from $(1,-1)$
That gives us $$(x-1)^2 + (-1)^2 =4$$ which implies $x=1pm sqrt 3 $
There are two circles with equations $$(x-1pm sqrt 3)^2 +(y+2)^2=4$$
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
add a comment |
up vote
0
down vote
up vote
0
down vote
Since the circle is tangent to the x-axis and passes through $(1,-1)$ with radius $2$ we know the center is below the x-axis and $2$ units away from it.
Thus the center is a point $ C(x,-2)$ which is $2$ unit apart from $(1,-1)$
That gives us $$(x-1)^2 + (-1)^2 =4$$ which implies $x=1pm sqrt 3 $
There are two circles with equations $$(x-1pm sqrt 3)^2 +(y+2)^2=4$$
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
Since the circle is tangent to the x-axis and passes through $(1,-1)$ with radius $2$ we know the center is below the x-axis and $2$ units away from it.
Thus the center is a point $ C(x,-2)$ which is $2$ unit apart from $(1,-1)$
That gives us $$(x-1)^2 + (-1)^2 =4$$ which implies $x=1pm sqrt 3 $
There are two circles with equations $$(x-1pm sqrt 3)^2 +(y+2)^2=4$$
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
answered Nov 18 at 11:52
Mohammad Riazi-Kermani
40.3k41958
40.3k41958
add a comment |
add a comment |
"my math teacher didnt teach this yet" A teacher shouldn't have to teach the exact method of solving a specific problem in order to expect his / her students to solve it. You have likely been taught all the tools you need: You know what the equation of a circle looks like, and you've been taught how to solve equations. In this problem you've been given information you can enter directly into that equation in order to solve and figure out the parts you haven't been told. Try it out, see how far you get, then let us know where you're stuck.
– Arthur
Nov 18 at 10:25
The circle also passes through $(x,0)$.
– Yadati Kiran
Nov 18 at 10:49
$(x-h)^2 + (y-k)^2=2^2 $ How to find $(h,k)?$
– Narasimham
Nov 18 at 18:17
Hint: the center has coordinates $(a,2)$ or $(a,-2)$.
– egreg
Nov 18 at 23:30