Here is the question which I am referring to:

A tangent is drawn to the circle $(x-a)^2+y^2=b^2$ and a perpendicular tangent to the circle $(x+a)^2+y^2=c^2$, find locus of their point of intersection.

What I did:

First I supposed the intersection of the perpendicular tangents to be $(h,k)$ and then from that point I found the equation of tangents to respective circles and after that I found slopes of each tangent using the condition that distance from center is equal to radius for a tangent and in the end. I multiplied the slopes of each tangent received from respective circles and set it equal to $-1$ because product of slopes of perpendicular lines is $-1$. I've found the locus but it doesn't seem as the answer. Can you tell me what mistake I made or is there any other way to approach this question?

Below are the images of my work:

Part 1 of work

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Part 2 of work

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The answer:

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  • $\begingroup$ There will in most cases be two tangents to the second circle that are perpendicular to the first tangent. How did you decide which one to pick? $\endgroup$
    – amd
    Commented Jul 18, 2016 at 17:10
  • $\begingroup$ @amd ya that also was the problem,I chosed the ones with positive irrational part slopes without any reason.Did you got some way to choose? $\endgroup$ Commented Jul 18, 2016 at 18:18

1 Answer 1


There're two pairs of tangents for the two circles are perpendicular. Let the contact points on the circle be $\begin{pmatrix} a+b\cos t \\ b\sin t \end{pmatrix}$, $\begin{pmatrix} -a-c\sin t \\ c\cos t \end{pmatrix}$, $\begin{pmatrix} a-b\cos t \\ -b\sin t \end{pmatrix}$ and $\begin{pmatrix} -a+c\sin t \\ -c\cos t \end{pmatrix}$.

The tangent from the first point: $$(a+b\cos t)x-a(x+a+b\cos t)+a^2+by\sin t=b^2$$

The tangent from the second point: $$(-a-c\sin t)x+a(x-a-c\sin t)+a^2+cy\cos t=c^2$$

On solving,

$$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a\cos 2t+b\cos t-c\sin t \\ a\sin 2t+b\sin t+c\cos t \end{pmatrix}$$

Similarly, we have another branch $$\begin{pmatrix} x \\ y \end{pmatrix} = \begin{pmatrix} a\cos 2t+b\cos t+c\sin t \\ a\sin 2t+b\sin t-c\cos t \end{pmatrix}$$

Each branch correspond to one diagonal of the rectangle.

enter image description here

Useful fact:

Equation of tangent for conics $ax^2+2hxy+by^2+2gx+2fy+c=0$ at the point $(x',y')$ is given by


  • $\begingroup$ interesting and cute(+1) $\endgroup$
    – G Cab
    Commented Jun 29, 2018 at 23:48
  • 1
    $\begingroup$ Implicitized those are $(x^2+y^2-a^2)^2-((b^2+c^2)(x^2+y^2+a^2)+2a(b^2-c^2)x\pm4a b c y)$, so it seems OP had one of the two branches. $\endgroup$ Commented Jun 30, 2018 at 3:34

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