= Solution
An <inversion in a circle> on the <Riemann sphere> is the anti-Möbius involution fixing that generalized circle pointwise. For a Euclidean circle with centre $a$ and radius $R>0$, it is
$$
\boxed{J_\Gamma(z)=a+\frac{R^2}{\overline z-\overline a},\qquad J_\Gamma(a)=\infty,\quad J_\Gamma(\infty)=a.}
$$
It reverses the argument relative to the centre and sends radius $r$ to $R^2/r$, so exactly the points of the circle are fixed. For a straight line, regarded as a circle through infinity, inversion is the ordinary reflection in that line. These formulas prove existence. For uniqueness, the composition of two anti-Möbius transformations fixing the same circle pointwise is a <Möbius transformation> fixing at least three distinct points, hence the identity. Therefore the two inversions agree.
Every inversion has the form $(a\overline z+b)/(c\overline z+d)$ with $ad-bc\ne0$. Composing two conjugate-linear fractional expressions cancels the <complex conjugations> and gives $(Az+B)/(Cz+D)$ with nonzero <determinant>. Induction proves that \b[an even number of inversions gives a <Möbius transformation>].
Conversely translations are compositions of reflections in parallel lines, rotations are compositions of reflections in two lines through their centre, and positive dilations are compositions of two concentric circle inversions: $J_RJ_1(z)=R^2z$. Thus every nonconstant affine map $z\mapsto az+b$ is an even composition. The map $z\mapsto1/z$ is $J_{|z|=1}$ composed with reflection in the real axis, also an even composition. If $c\ne0$, factor a general <Möbius transformation> as
$$
\frac{az+b}{cz+d}=\frac ac+\frac{bc-ad}{c^2}\frac1{z+d/c}.
$$
This is a composition of affine maps and $1/z$. If $c=0$, it is already affine. \b[Every <Möbius transformation> is therefore a composition of an even number of inversions], with the identity represented by zero inversions or by any inversion repeated twice.
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