Prove conjecture about circle pairing centers I'm desperately trying to prove a statement which I need for some other results in my bachelor final project (I already spent several hours on this and it's driving me crazy). I would be very happy if you could help me, and if you can you will get credited in my bachelor final project report. Here are some definitions, theorems and terminology.
Definition: a circle pairing $f$ is a Möbius transformation which maps the exterior of a circle $C_F$ to the interior of a circle $C_f$, and it maps the interior of $C_F$ to the exterior of $C_f$ (and from this follows that it maps $C_F$ onto $C_f$).
Terminology: for a circle pairing $f$ as above, the circle $C_f$ has center $C_1$, and the circle $C_F$ has center $C_2$.
Theorem: for any given triple of points there is a unique Möbius transformation which maps these three points to any other given triple of points.
Theorem: Möbius transformations map circles and lines to circles or lines.
The statement I want to prove:
For every circle pairing $f$ the circles $C_F$ and $C_f$ which are paired as above can be chosen such that $f$ maps the point at infinity to $C_1$, and maps $C_2$ to the point at infinity. Assume that the point at infinity lies in the exterior of both circles which are paired.
My question is if you can prove this, either by a totally different way (maybe this is trivial if you look at it from some different perspective or something), or finish my incomplete proof below. Of course you would also answer my question if you can find a counterexample but I doubt you will be able to.
Here is what I have so far:
(Incomplete) proof:
Fix a circle pairing $f$ such that the point at infinity is in neither circle which is paired. In other words, fix a Möbius transformation $f$ such that there exist 2 circles $J_1, J_2$ such that the point at infinity is in the exterior of both, and such that the exterior of $J_2$ is mapped to the interior of $J_1$, and the interior of $J_2$ is mapped to the exterior of $J_1$. Say $f$ has coefficients $a,b,c,d$. We have that $f$ maps the point at infinity to $a/c$, as is the case for all Möbius transformations.  Denote the inverse of $f$ as $F$, which clearly has coefficients $d,-b,-c,a$. We have that $F$ maps the point at infinity to $-d/c$. This implies that $f$ maps $-d/c$ to the point at infinity. Consider the line through $a/c$ and $-d/c$, and choose some complex number $H$ which lies on this line, between $a/c$ and $-d/c$ (so formally such that the distance from $H$ to both $a/c$ and $-d/c$ is strictly smaller than the distance between $a/c$ and $-d/c$). Given that we already know that $inf->a/c, -d/c->inf$ and $H=/=a/c,-d/c$, we have that $H$ is mapped to some complex number $J$ (Side note: by a theorem above we have that all information about the circle pairing has to be contained in the fact that inf->$a/c$, $-d/c$->inf, $H$->$J$, so $H$ and $J$ would somehow have to capture information about circles around $a/c$ and $-d/c$). Define the circle $C_F$ through $H$ with center $-d/c$. We have that $C_F$ is mapped to some circle $C_f$ through $J$ (by one of the theorems above it is mapped to a line or circle, but because $-d/c$ is not on the circle it can only be mapped to a circle, and because $H->J$ it must be a circle through $J$; FYI a line is a circle through the point at infinity). If we can show that this image circle always has $a/c$ as center we are done as by some theorem in topology it follows that then $f$ is a circle pairing where the circles which are paired are $C_F$ and $C_f$.
Possible leads which might help with finishing the proof (from huge leads to less promising leads):

*

*Maybe we can argue using the inverse map $F$, which is also a circle pairing, hence it should behave similarly to $f$. As $f$ maps $C_F$ to some circle $C_f$ through $J$ we have that $F$ maps $C_f$ to $C_F$. Furthermore we have because $F$ is also a circle pairing, that we can consider the circle $C_g$ through $J$ (confusing notation maybe) with as center $a/c$ and say that it is mapped to some circle through $H$. Now we have to show that $C_g=C_f$. Maybe this reasoning is even better as it could imply something for $f$ directly, but I'm not sure.

*Something related to symmetry. If it somehow follows that by symmetry there must be some other point on $C_F$ which is mapped to the reflection of $J$ in the line through $a/c$ and $-d/c$, and another point on $C_F$ is mapped to some other point on this line, then maybe that could help.

*Something related to the side note in my incomplete proof above. Maybe there is some shortcut by using this in combination with some other theorem I don't know about.

*It should be possible to argue that $a/c$ must lie inside $C_f$ by the fact that we are considering a circle pairing. I.e. use the fact that we know that at least some circles are paired, and because the point at infinity is in neither is is mapped to the interior of the other circle.

*Something related to the distance between $H$ and the other intersection point of the line through $a/c$ and $-d/c$ which is on $C_F$. This might give a clue on the diameter of the image circle.

*The line through $a/c$, $-d/c$ and $H$ is mapped to a line through $J$ and the image of $a/c$. For the image of $a/c$ we know that it is mapped to some number $W$ which has smaller distance to the attracting fixed point of $f$.

*The only line which can be preserved is the line through $a/c$ and $-d/c$ (as the point at infinity lies on every line, is mapped to $a/c$ and $-d/c$ is mapped to the point at infinity).

*If the attracting fixed point lies on the line through $a/c$ and $-d/c$ then it is mapped to the line through $J$ and that fixed point.

Thanks in advance if you are willing to help :)
 A: For a Mobius transformation $f$ to turn a circle inside-out, it cannot fix $\infty$. (The set of transformations which fix $\infty$ is precisely the set of affine transformations $z\mapsto az+b$.)
Your statement is then that for any transformation $f$ not fixing $\infty$ there exist circles $A$ and $B$ with centers $\alpha$ and $\beta$ so that $f$ not only maps the interior/exterior of $A$ to the exterior/interior of $B$ (respectively, and thus sends $A$ to $B$), but more precisely sends $A$'s center to $\infty$ and $\infty$ to $B$'s center.
Suppose $f(z)=\frac{az+b}{cz+d}$. The centers of $A$ and $B$ are thus determined by $f$: the center of $A$ is given by $\alpha=f^{-1}(\infty)=-\frac{d}{c}$ (since $f^{-1}(z)=\frac{\phantom{-}dz-b}{-cz+a}$) and the center of $B$ is given by $\beta=f(\infty)=\frac{a}{c}$.
Define, then, $g(z)=f(z+\alpha)-\beta$. Then $g$ is a transformation that swaps $0$ and $\infty$. Indeed, $g$ is a circle pairing for the translated circles $A'=A-\alpha$ and $B'=B-\beta$ (both centered at $0$) if and only if $f$ is a circle pairing for the circles $A$ and $B$ centered at $\alpha$ and $\beta$. The transformations $g$ which swap $0\leftrightarrow\infty$ are all of the form $f(z)=w/z$ (where $w\ne0,\infty$). Indeed, $g$ can be considered a circle pairing with $A'$ an arbitrary circle centered at $0$ (and $B'=g(A'$), then, which has radius $|w|/r$ if $A'$ has radius $r$). This in turn means $f$ must be a circle pairing between $A$ and $B=f(A)$ for any circle $A$ centered at $\alpha=f^{-1}(\infty)$ (which also means $B=f(A)$ is automatically centered at $\beta=f(\infty)$).
