Past exam of the mathematics course of the University of Cambridge / 2008 / ii / Paper 1 / 11G / Solution

inversion in a circle with center $c$ and radius $r>0$ is $I(z)=c+r^2/(\overline{z}-\overline{c})$, with $c$ and infinity interchanged. It fixes the circle pointwise and reverses complex orientation. Each inversion is fractional-linear in $\overline{z}$, so composing two eliminates the conjugation and gives $(az+b)/(cz+d)$ with nonzero determinant: a Möbius transformation.

Translate and rotate two distinct centers to $0,d$, where $d>0$, and let the radii be $r,s$. The composition of inversion in the first circle followed by the second isIts fixed points solve $d{z}^2-(d^2+r^2-s^2)z+d{r}^2=0$. The discriminant isDisjoint circle boundaries require either $d>r+s$ or $d<|r-s|$, so both factors have the same nonzero sign and $\Delta >0$. There are two distinct fixed points on the line of centers. If the circles are concentric, their radii differ and the composition is $z\mapsto (s^2/r^2)z$, with exactly the two fixed points $0,\infty$. Thus the composition has two distinct fixed points, also for nested disjoint circles. Reversing the order inverts the Möbius transformation and leaves its fixed points unchanged.