how to find inverse point in a complex plane

Question

How to find the inverse point of the point z=a with respect to the circle $|z-c|=r$ (where c is the origin and r the radius) ?

$c+\frac{r^{2}}{a-c}$

this is the answer given in the book...how do we get it

Answer 1

Start by noting that the inverse point of $a$ wrt the unit circle centre the origin is $1/\,\overline{a}$. Hence the inverse point of $a$ a circle radius $r$ centre the origin is $\frac{r^2}{\overline{a}}$. Now we have to translate that to a circle centre $c$. Relative to $c$, the point $a$ is $a-c$, so we want $c+\frac{r^2}{\overline{a-c}}$.

Answer 2

If $z_0,p,q$ are collinear points, where $z_0$ is center of circle with radius $k$, $q$ lies outside the circle, and $p$ is inverse point of $q$ which lies inside the circle, then $$p=\frac{z_0+k^2(q-z_0)}{|q-z_0|^2}$$

$$P =\frac{z_0+k^2(q-z_0)}{(q-z_0)(q-Z_0)}$$

$$P=\frac{z_0+k^2}{q-z_0}$$