Let $H$ denote the upper half-plane model of hyperbolic space. If $L$ is the hyperbolic line in $H$ given by a Euclidean semicircle with centre $a\in \mathbb{R}$ and radius $r >0$, show that reflection in the line $L$ is given by the formula $R_{L}(z)=a+ {r^2 \over{\overline{z} -a}} $.
[Math] Reflection in a hyperbolic line formula
geometryhyperbolic-geometry
Related Solutions
Making my comment more explicit ...
Writing $|\cdot|$ for Euclidean distance, and $|\cdot|^\star$ for hyperbolic distance, we have a relatively simple relation for distances from the origin:
$$|OX|^\star = \log\frac{1 + |OX|}{1 - |OX|} = 2 \operatorname{atanh}|OX| \qquad\qquad |OX| = \tanh\frac{\;|OX|^\star}{2} \tag{$\star$}$$
Let the diameter of the target circle meet $\overleftrightarrow{OA}$ at $P$ and $Q$, and define $a := |OA|$, $p := |OP|$, $q := |OQ|$, with $a^\star$, $p^\star$, $q^\star$ their hyperbolic counterparts. Let $r^\star$ be the target circle's hyperbolic radius.
We may assume $p \geq a$ (one of the diameter's endpoints must be on the "far side" of center $A$), so that
$$p^\star = a^\star + r^\star \quad\to\quad p = \tanh\frac{a^\star + r^\star}{2} = \frac{(1+a)\exp r^\star - (1 - a)}{(1+a)\exp r^\star + ( 1 - a)} \tag{1}$$
For the other endpoint, $Q$, an ambiguity arises based on whether the origin lies outside or inside the circle, but we have
$$q^\star = \pm ( a^\star - r^\star ) \quad\to\quad q = \pm \tanh\frac{a^\star - r^\star}{2} = \pm \frac{(1+a) - ( 1 - a )\exp r^\star}{(1+a) + (1-a)\exp r^\star} \tag{2}$$
where "$\pm$" is "$-$" for $O$ inside the circle, and "$+$" otherwise. (If you like, you can absorb the sign into the distances $q$ and $q^\star$, so that they are negative when $\overrightarrow{OA}$ and $\overrightarrow{OQ}$ point in opposite directions, and positive otherwise.)
With the endpoints of the target circle's diameter known, determining the Euclidean center and Euclidean radius is straightforward. $\square$
Note. If $R$ is such that $|OR|^\star = r^\star$, and if we define $r := |OR|$, then $(1)$ and $(2)$ become:
$$p = \frac{a + r}{1 + a r} \qquad\qquad q = \pm \frac{a - r}{1 - a r} \tag{3}$$
Here's a proof which takes advantage of symmetry.
The key fact you need is that Möbius transformations of $\mathbb{C} \cup \{\infty\}$ preserve the set $\{\text{circles}\} \cup \{\text{lines}\}$; this is not hard to prove directly.
It follows that group of Möbius transformations of the upper half plane preserves the set of Euclidean circles in the upper half plane (the only lines in the upper half plane are horizontal and transform to themselves or to Euclidean circles tangent to the real line, none of which is a Euclidean circle entirely contained in the upper half plane). Also, hyperbolic circles are preserved under that action, since that action is the same as the group of orientation preserving isometries of the hyperbolic metric on the upper half plane.
There exists a Möbius transformation that transforms the Poincare disc to the upper half plane, this transformation takes Euclidean circles to Euclidean circles (there are no lines in the Poincare disc), and it takes hyperbolic circles to hyperbolic circles (it is an isometry between the hyperbolic metrics on the Poincare disc and the upper half plane).
So, we have reduced the problem to showing that in the Poincare disc with the hyperbolic metric, hyperbolic circles are the same as Euclidean circles.
The group of Möbius transformations of the Poincare disc equals the group of orientation preserving isometries of the hyperbolic metric on the Poincare disc, and this action is transitive on points, hence for each $r>0$ action is transitive on the set of hyperbolic circles of hyperbolic radius $r$. Thus, it suffices to find, for each $r>0$, a single example of a hyperbolic circle of hyperbolic radius $r$ which is simultaneously a Euclidean circle: and there is evidently a Euclidean circle centered on the origin which is a circle of hyperbolic radius $r$.
Best Answer
Hint: $\operatorname{Isom}(\mathbb{H}) \cong SL_2(\mathbb{R})$ acting on $\mathbb{H}$ by Möbius transformations. Identify the Möbius transformation that preserves $L$ and swaps $a$ and $\infty$.