Locus of intersection point of perpendicular tangents 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

Part 2 of work

The answer:

 A: 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.


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

