= 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 is
$$
T(z)=d+\frac{s^2z}{r^2-dz}.
$$
Its <fixed points> solve $dz^2-(d^2+r^2-s^2)z+dr^2=0$. The <discriminant> is
$$
\Delta=(d^2-(r+s)^2)(d^2-(r-s)^2).
$$
Disjoint 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 \b[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.
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