Question

The chords of contact of tangents drawn from a variable point to two fixed circles are perpendicular to each other. Prove that its locus is a circle whose centre is at the mid-point of the line joining the centres of two given circles.

Answer 1

Let the circle by $x^2+y^2=a^2$ and
$x^2+y^2+2gx+2fy+c=0$
If the variable point $P$ be $(h,k)$, then its chord of contact w.r.t. to given circles are
$hx+ky-a^2=0.....\left(1\right)$
$hx+ky+g(x+h)+f(y+k)+c=0....\left(2\right)$
The lines $\left(1\right)$ and $\left(2\right)$ are perpendicular
$\therefore a_1{a}_2+b_1{b}_2=0$
or $h(h+g)+k(k+f)=0$
$\therefore$ locus of $(h,k)$ is
$x(x+g)+y(y+f)=0$
Above represents a circle whose diameter is $(0,0)$ and $(-g,-f)$ and hence its centre is $(-g/2,-f/2)$
i.e. the mid-point of centres of given circles.