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I am given that $P(-3,-1)$ and $Q(5,3)$ are points of the circle. Also, the line $L:0=x+2y-13$ is tangent to said circle. The objective is to find the equation of the circle.

I thought of a way for solving this, but it doesn't seem to be the best one. The option was to find the points equidistant to $P$ and $Q$. Afterwards, using any point $(x,y)$ of the line of points equidistant to $P$ and $Q$, the next step would be to make the distance from $(x,y)$ to $L$ the same as to the distance to $P$ or $Q$.

Equations: $$ 2a + b = 3 \\ 5a^2-10a+25 = r^2\\ \sqrt{5}r = a+2b-13 $$

The solution for posterity

OFRBG
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  • I'm not understanding where the tangent line comes into your approach. – dfeuer Oct 13 '13 at 19:02
  • For anyone else struggling to read the diagram, $P$ is labelled A, $Q$ is labelled B, and $L$ is the line CD – Eric Jan 22 '20 at 15:56

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HINT:

If $(a,b)$ is the center of the required circle

we have $$(a+3)^2+(b+1)^2=(a-5)^2+(b-3)^2=r^2$$ where $r$ is the radius

Now, the perpendicular distance of the tangent form the center $(a,b)$ is again equal to the radius

Do you know how to calculate the perpendicular distance?

lab bhattacharjee
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  • So I get the relation $2a + b = 3$, and either of the other two equations above. There is also the formula of distance from a line to a point, but substituting, it looks messy. Am I doing something wrong here? – OFRBG Oct 13 '13 at 19:22
  • @Fiire, the square of the perpendicular distance $=(a+3)^2+{3-2a+1}^2$ – lab bhattacharjee Oct 13 '13 at 19:24
  • Sorry about this, but from where does that come? I know the one that I recently edited in above. – OFRBG Oct 13 '13 at 19:29
  • Never mind, I see what you did there. – OFRBG Oct 13 '13 at 19:31
  • @Fiire, I have just replaced $b=3-2a$ – lab bhattacharjee Oct 13 '13 at 19:31
  • Still, I get a quadratic, right? That gives $r^2$ in terms of $a$ in a quadratic form. The third equation would be the one of the distance from the tangent line to the center of the circle, but that only makes it worse. Is this true? – OFRBG Oct 13 '13 at 19:41
  • @Fiire, the square of radius $(a+3)^2+(3−2a+1)^2$ and the square of the perpendicular distance $=\left(\frac{a+2b-13}{\sqrt5}\right)^2=\left(\frac{a+2(3-2a)-13}{\sqrt5}\right)^2=\left(\frac{3a+7}{\sqrt5}\right)^2$ – lab bhattacharjee Oct 13 '13 at 19:44
  • Thanks for your help. I was doing my arithmetics wrong, so I got stuck. Without your help I would still be scratching my head.The solution is up there. – OFRBG Oct 13 '13 at 20:16