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Let $\hat{\mathbb{C}}=\mathbb{C} \cup \left\{{ \infty}\right\}$ be the extended complex plane, let's define a circle in $\hat{\mathbb{C}}$ as a set $K$ which is whether an ordinary circle in $\mathbb{C}$ or a line $L \subset \mathbb{C}$ together with $\left\{{ \infty}\right\}$. Denote by $C$ the set of all the circles in $\hat{\mathbb{C}}$.

Once said that, let $z_1,z_2 \in \mathbb{C}$ be two distinct points and let $$ C_{z_1,z_2}=\left\{{K \in C: z_1,z_2 \in K}\right\} $$ Also, consider the set $$ C_{z_1,z_2}^{\perp}=\left\{{K' \in C: K' \perp K \textrm{ for all } K \in C_{z_1,z_2} }\right\} $$ Is it true that if $K' \in C_{z_1,z_2}^{\perp} $ then there exist $r >0$ such that $K'=\left\{{z: \vert z-z_1 \vert}=r \vert z-z_2 \vert \right\}$? If so, how can I prove it? I am asking this because I am asked to prove that if $T$ is a Möbius transformation with two distinct fixed points $z_1,z_2$ and there exist a $\mu>0$ such that $$ \frac{T(z)-z_1}{T(z)-z_2}=\mu \frac{z-z_1}{z-z_2} $$ for all $ z\in \mathbb{C}$, then $T$ "exchanges" the circles of $C_{z_1,z_2}^{\perp}$. Therefore, if the answer to my question is affirmative it's more or less clear how to prove it, otherwise I don't know how to proceed.

In advance thank you very much.

ferolimen
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    See Apollonius circles (here](https://math.stackexchange.com/q/520469/305862) for example. – Jean Marie Apr 25 '23 at 21:36
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    as to why the thing is possible: take a round sphere of an appropriate radius, place the North p[ole at $z_1$ and the South pole at $z_2$ take a line through the center of the sphere that is also perpendicular to the plane; it intersects the sphere in two points, call one P. The sphere has all latitude circles and longitude circles, umm, built in. Finally, stereographic projection out of the point P send the latitude and longitude circles to circles in the plane, while the original equator is sent to a line. The image curves are circles and pairwise orthogonal – Will Jagy Apr 25 '23 at 22:34

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If you only want to prove that $T$ exchanges the circles in $C_{z_1,z_2}^{\perp}$ (assuming that $z_1,z_2$ are two distinct fixed points) then you don't need such a characterization of the circles in $C_{z_1,z_2}^{\perp}$. Indeed, let $S$ be any Möbius transformation such that $S(z_1)=0$ and $S(z_2)=\infty$, then we can write $$ S \circ T \circ S^{-1}=\mu I $$ where $I$ is the identity map, that is, $I(z)=z$. Now, it is worth noticing that $C_{z_1,z_2}^{\perp}=S^{-1}(C_{0,\infty}^{\perp})$. Therefore, if $K \in C_{z_1,z_2}^{\perp}$ then there exist $K' \in C_{0,\infty}^{\perp} $ such that $C=S^{-1}(K')$ and thus $$ T(K)=(S^{-1} \circ \mu I)(K') $$ Now, since $K'$ must be orthogonal to each line that passes through the origin it follows that $K'$ is a circle concentric to the origin and consequently, since $\mu>0$ we have that $\mu I (K')$ is again a circle concentric to the origin (not equal to $K'$) and hence $T(K) \in S^{-1}(C_{0\infty}^{\perp})=C_{z_1,z_2}^{\perp}$ but $T(K) \neq K$ which means that $T$ exchanges the circles in $C_{z_1,z_2}^{\perp}$. Regarding your first question, I am not sure if each circle in $C_{z_1,z_2}^{\perp}$ is necessary an apollonius circle, that is, a set of the form $\left\{{z: \vert z-z_1 \vert = r \vert z-z_2 \vert }\right\}$ but it will be a good thing to come up with a proof or a counterexample though.

SCarlsen
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