# Equation of a circle from given conditions

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## Homework Statement

Equation of the circle passing through the point (1,2) and (3,4) and touching the line 3x+y-3=0 is?

## Homework Equations

x^2+y^2+2gx+2fy+c=0...(1)
(-g,-f)=center of circle

## The Attempt at a Solution

Putting (1,2) and (3,4) in equation 1 we get 5+2g+4f+c=0; 25+6g+8f+c=0.
Now, line joining the two points will be perpendicular to the line joining center and midpoint of that line (chord perpendicular to radius). Say (h,k) is center, slope joining the two points is 1 so slope of radius through midpoint is -1 (perpendicular lines), midpoint of chord is (2,3); equating -1 to slope of (h,k) and (2,3) gives us k+h=5- but h= -g and k= -f; so -g-f=5 Solving these three equations gives c=40, f= -35/2 and g= 25/2 which is the wrong circle. I know there are other ways to solve this but I want to know why this method is not working in particular- I double checked all the calculations and I can't figure out anything wrong with my logic, Thank you for your help

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It is unclear to me what you are trying to do with this ”method”. Why are you creating a line from the centre to the midpoint? You have already used that the two points need to be on the circle and there are an infinite number of circles satisfying this. You cannot squeeze more information out of those two points. You need to use the third requirement.

Delta2
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so -g-f=5 Solving these three equations gives c=40, f= -35/2 and g= 25/2
Also note that your equation in bold here is not a new equation. You can get it by just using your previous two so there is no new information. Your equation system therefore does not have a unique solution (two equations for three variables) and you need to use the extra information provided.

Gold Member
It is unclear to me what you are trying to do with this ”method”. Why are you creating a line from the centre to the midpoint? You have already used that the two points need to be on the circle and there are an infinite number of circles satisfying this. You cannot squeeze more information out of those two points. You need to use the third requirement.
Oh, right. I just realized that- sort of like writing a third KVL equation which is the same. To use the third condition I'd have to put the radius=distance of line from center which is lengthy and prone to mistakes in an exam. I was hoping to find a shorter method but I suppose this is the only way to do it...

Delta2
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Also note that your equation in bold here is not a new equation. You can get it by just using your previous two so there is no new information. Your equation system therefore does not have a unique solution (two equations for three variables) and you need to use the extra information provided.
Thank you very much for your help :D

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$$(x-a)^2+(y-(5-a))^2=r^2$$ Minimise for tangent: $$2(x-a)+2(y-(5-a))\frac{dy}{dx}=0$$ with $$y=3-3x \Rightarrow \frac{dy}{dx}=-3$$ Interesting problem - you end up with two values of a and hence two circles satisfying the given conditions.

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$$(x-a)^2+(y-(5-a))^2=r^2$$ Minimise for tangent: $$2(x-a)+2(y-(5-a))\frac{dy}{dx}=0$$ with $$y=3-3x \Rightarrow \frac{dy}{dx}=-3$$ Interesting problem - you end up with two values of a and hence two circles satisfying the given conditions.
how did you obtain coordinates of center as (a,5-a)?

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how did you obtain coordinates of center as (a,5-a)?
From equation of perpendicular bisector of the line drawn between the two given points.

SammyS
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From equation of perpendicular bisector of the line drawn between the two given points.
ohh, really good solution- how did you think of this?

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Other approaches seemed to be heading for complications so I tried to keep it simple! I wasn't quite sure how to use the information about the tangent line until I realized the problem was essentially one of minimising distance between point (the circle centre) and line (y=3-3x).

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Other approaches seemed to be heading for complications so I tried to keep it simple! I wasn't quite sure how to use the information about the tangent line until I realized the problem was essentially one of minimising distance between point (the circle centre) and line (y=3-3x).
Great! Thank you very much for your help.

Homework Helper
ohh, really good solution- how did you think of this?
A pleasure. Thanks for your kind compliment - the problem was certainly a little different from 'run of the mill' exercises in analytic geometry.