Let $A$ and $B$ be disjoint circles in $\mathbb{C}$. Prove that there is a Möbius transformation which takes $A$ and $B$ to two concentric circles.

A collection of circles $X_{i} \subset \mathbb{C}, 0 \leqslant i \leqslant n-1$, for which

1. $X_{i}$ is tangent to $A, B$ and $X_{i+1}$, where indices are $\bmod n$;

2. the circles are disjoint away from tangency points;

is called a constellation on $(A, B)$. Prove that for any $n \geqslant 2$ there is some pair $(A, B)$ and a constellation on $(A, B)$ made up of precisely $n$ circles. Draw a picture illustrating your answer.

Given a constellation on $(A, B)$, prove that the tangency points $X_{i} \cap X_{i+1}$ for $0 \leqslant i \leqslant n-1$ all lie on a circle. Moreover, prove that if we take any other circle $Y_{0}$ tangent to $A$ and $B$, and then construct $Y_{i}$ for $i \geqslant 1$ inductively so that $Y_{i}$ is tangent to $A, B$ and $Y_{i-1}$, then we will have $Y_{n}=Y_{0}$, i.e. the chain of circles will again close up to form a constellation.