Prove conjecture about classification of circle pairings

Question

I need help proving or disproving a conjecture related to circle pairings, which I'm trying to prove for my bachelor final project. I first present some needed terminology and context.

A circle pairing of circles $C$ and $D$ is a Möbius transformation which maps $D$ to $C$, maps the exterior of $D$ to the interior of $C$, and maps the interior of $D$ to the exterior of $C$.

The circles which are paired are not unique. I'm trying to classify circle pairings based on the absolute value of the trace of the corresponding matrices (the group of Möbius transformations is isomorphic to the group of 2x2 invertible matrices modulo plusminus). I have already found the following results:

Theorem 1: the most general way any circle pairing that pairs $C(x_1, r_1)$ and $D(x_2, r_2)$ can be written as is $$g(z)=x_1+r_1\cfrac{\bar{v}(z-x_2)+r_2 \bar{u}}{u(z-x_2)+r_2 v},$$ where $u,v\in \mathbb{C}:|u|^2-|v|^2=1$, where $x_1, x_2$ are the centers and $r_1, r_2$ are the radii of $C$ and $D$.

Lemma: the most general way any circle pairing $f$ that pairs circles $C(x_1, r_1)$ and $D(x_2, r_2)$ such that $f(x_2)=\infty$ and $f(\infty)=x_1$, can be written as is $$x_1 +\cfrac{(H-x_2)(J-x_1)}{z-x_2},$$ where $H\in D$ and $J\in C$ (so $|H-x_2|=r_2$ and $|J-x_1|=r_1$).

Theorem 2: for every circle pairing $g$ given as in Theorem 1, which maps $\tilde{x}_2\mapsto \infty$ and $\infty \mapsto \tilde{x}_1$ there exist circles $C(\tilde{x}_1, r_1/|u|)$ and $D(\tilde{x}_2,r_2/|u|)$ such that $g$ pairs $C$ and $D$. In other words, every circle pairing as in Theorem 1, can be written in the form of the lemma above.

Proof: We denote $\tilde{x}_1=x_1+r_1 \bar{v}/u$, $\tilde{x}_2=x_2-r_2 v/u$, $\tilde{r}_2=r_2/|u|=|H-\tilde{x}_2|$ and $\tilde{r}_1=r_1/|u|=|J-\tilde{x}_1|$, where $H=x_2+\tilde{r}_2$ and $J=\tilde{x}_1+\tilde{r}_1$. We get $$\begin{equation*} \begin{split} \tilde{x}_1 +\cfrac{(H-\tilde{x}_2)(J-\tilde{x}_1)}{z-\tilde{x}_2} & = x_1+r_1 \frac{\bar{v}}{u}+\cfrac{r_1 r_2 / u^2 }{(z-x_2)+r_2 v/u} \\ & = x_1+r_1(\frac{\bar{v}}{u}\frac{1/u [(z-x_2)+r_2 v/u]}{1/u [(z-x_2)+r_2 v/u]}+\cfrac{r_2/u^2}{(z-x_2)+r_2 v/u}) \\ & = x_1+r_1\cfrac{\bar{v}(z-x_2)+\bar{v} r_2 v/u+r_2/u}{u(z-x_2)+r_2 v } \\ & = x_1+r_1 \cfrac{\bar{v}(z-x_2)+r_2(\bar{v}v+1)/u}{u(z-x_2)+r_2 v} \\ & = x_1+r_1\cfrac{\bar{v}(z-x_2)+r_2 \bar{u}}{u(z-x_2)+r_2 v}. \\ \end{split} \end{equation*}$$

The absolute value of the trace of a circle pairing $f$ as in Theorem 1 is $|Tr f|=\cfrac{|\tilde{x}_2-\tilde{x}_1|}{\sqrt{r_1 r_2 /|u|^2}}$, where $f(\tilde{x}_2)=\infty$ and $f(\infty)=\tilde{x}_1$ (so $\tilde{x}_1=x_1+r_1 \bar{v}/u$ and $\tilde{x}_2=x_2-r_2 v/u$). I have shown that if $|Tr f|>2$, then the radii of the circles $C$ and $D$ as given the existence of by Theorem 2 can be chosen such that $C$ and $D$ are disjoint (in short, if the $R_1=r_1/|u|$ and $ R_2=r_2/|u|$ are the radii of the paired circles then it can be shown that $sR_2$, $R_1/s$, where $s>0$, are also paired by $f$). If $|Tr f|=2$, then these circles can be chosen to touch (i.e. intersect in precisely one point), but never be disjoint. If $|Tr f|<2$, then these circles can only be chosen to overlap. The reason this is true is that the value for $s$ for which these circles are the 'most disjoint' is such that the new radii $\hat{R_1}$ and $\hat{R_2}$ are such that $\hat{R_1}=\hat{R_2}=\sqrt{R_1 R_2}$. Circles are disjoint if $|x_1-x_2|>\text{sum of possible radii} \geq 2\sqrt{R_1 R_2}$, from which these conditions can be derived.

We call a circle pairing $f$ disjoint if $|Tr f|>2$, touching if $|Tr f|=2$ and overlapping if $|Tr f|<2$.

Conjecture: if a circle pairing $g$ is touching, then there exist no disjoint circles $C$ and $D$ such that $g$ pairs these. If a circle pairing $g$ is overlapping, then there exist no disjoint and no touching circles $C$ and $D$ such that $g$ pairs these. In other words, the freedom of how disjoint the circles as given by Theorem 2 can be chosen, is such that it is the 'most disjoint' case of all possible circles that can be chosen to be paired.

Theoretically, for some circle pairing $g$ as given by Theorem 1, the points which are mapped from and to $\infty$ have to just be somewhere in the interiors of $C(x_1,r_1)$ and $D(x_2,r_2)$ which are paired by $g$. If they are the centers $x_1$ and $x_2$, then the classification of the conjecture is trivially true, but it is not clear what happens if they are not.

Some things which may help to prove the conjecture which I have already considered but have not succeeded with:

1. Note that the product of the radii of the circles as given by Theorem 2, namely $r_1 r_2 /|u|^2$, is smaller than that of the original radii $r_1$ and $r_2$ as given by Theorem 1, as $|u|>1$.
2. Note that each of the fixed point(s) $f_{1,2}=1/2[\tilde{x}_1+\tilde{x}_2\pm \sqrt{4 r_1 r_2/u^2+(\tilde{x}_1-\tilde{x}_2)^2}]$, where again $\tilde{x}_1=x_1+r_1 \bar{v}/u$ and $\tilde{x}_2=x_2-r_2 v/u$, must either be in the interior of PRECISELY one of the circles that are paired, or be the intersection point(s) of these circles (this follows from the definition of circle pairings).
3. Maybe the equation of $g$ as given by Theorem 1 can be rewritten such that it equals itself, but where the values of $x_1, x_2, r_1, r_2, u$, and $v$ are different. This should theoretically be possible for every pair of circles which $g$ can pair. Maybe this could give insight into the freedom of possible circles which can be chosen to be paired, and why hypothetically the case where the centers are the points that map from and to $\infty$, would be the 'most disjoint' case.
4. Maybe $|Tr f|$ can be related directly to the 'disjointness' of circles which can be chosen to be paired by $g$, similarly to how I did it already for the case for which the centers are the points which are mapped from and to $\infty$. In other words, maybe the logic itself I used to come up with the classification for the simple case, can be generalized to the general case.

Hopefully anyone can find a proof or a counterexample. Thanks in advance!

Answer 1

The conjecture is false :( A counterexample is the circle pairing given by $x_1=0, x_2=2, r_1=r_2=0.99999, u=\sqrt{10001}, v=100$. The points that are mapped from and to the point at infinity are $x_1+r_1 \bar{v}/u=100/\sqrt{10001}\approx 0.99$ and $x_2-r_2 v/u=2-100/\sqrt{10001}\approx 1.01$. This is a circle pairing of disjoint circles $C(0,0.99999)$ and $D(2,0.99999)$, but the absolute value of the trace is $\approx |1.01-0.99|/\sqrt{1}\approx 0 < 2$, which means that the circle pairing is overlapping. However, by construction there exist disjoint circles which are paired by the circle pairing.