Question

From points on a given circle tangents are drawn to another circle. Prove that the locus of the mid-points of the chord of contact is also a circle.

Answer 1

Let $(p,q)$ be a point on a given circle.
$x^2+y^2+2gx+2fy+c=0$ so that
$p^2+q^2+2gp+2fq+c=0....\left(1\right)$
Let tangents be drawn to the circle $x^2+y^2=a^2$ from $P$ and mid-point of chord of contact $AB$ be $(h,k)$.
$AB$ is $px+qy=a^2$ as chord of contact.
$AB$ is $hx+ky=h^2+k^2$, by $T=S_1$
Comparing,
$\frac{p}{h}=\frac{q}{k}=\frac{a^2}{h^2+k^2}$
$\therefore p=\frac{a^{2}h}{h^2+k^2},q=\frac{a^{2}k}{h^2+k^2}$
Put these values of $p$ and $q$ in $\left(1\right)$ and generalize $(p,q)$ and you get a circle.