= Solution
For a Euclidean circle with centre $c$ and radius $r$, two distinct finite points are inverse when they lie on the same ray from $c$ and their distances from $c$ have product $r^2$. Points on the circle are fixed, and $c$ and infinity are inverse to each other. In complex coordinates this uniquely defines <inversion in a circle> by
$$
I_\Gamma(z)=c+\frac{r^2}{\overline z-\overline c}.
$$
Uniqueness follows from the ray and product conditions, including the specified centre–infinity pair; the formula also gives $I_\Gamma^2=\operatorname{id}$. A circle through infinity is a straight line in the complex plane; its inversion is reflection in that line, fixing infinity. For a line through $c$ with direction angle $\theta$, its formula is $c+e^{2i\theta}\overline{z-c}$.
Thus every inversion is an anti-<Möbius transformation>, of the form $(a\overline z+b)/(c\overline z+d)$ with nonzero determinant. If $J(z)=\overline z$, write such an inversion as $M\circ J$ with $M$ a <Möbius transformation>. Since $J\circ M\circ J$ is the Möbius transformation with conjugated coefficients, the product of two inversions is $M_1\circ(J\circ M_2\circ J)$, hence Möbius. Pairing consecutive inversions proves \b[every composition of an even number of inversions is a Möbius transformation].
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