Q:

Hyperbolic inversions are transitive on unit vectors at $x \in D$

Consider the Poincaré model in which the hyperbolic plane is the interior of a disk $D$, and a point $x$ in it with two vectors $v$ and $w$ of the same length attached. The reflection with respect to any circle $C$ passing through $x$ fixes $x$ and is orthogonal (it maps the normal and tangent vectors to $C$ at $x$ to themselves respectively). But how can I find a circle such that the reflection with respect to it will take $v$ to $w$?

A:

Here are two sketches (that should at least get you started even if they're not exactly what you're seeking):

(Geometric) Let $c$ be a non-zero vector at $x$ that bisects the hyperbolic angle between $v$ and $w$, and let $C$ be the hyperbolic line tangent to $c$ at $x$. Euclidean inversion in $C$ (i.e., hyperbolic reflection across $C$) exchanges $v$ and $w$.
(Computational) Pick a hyperbolic isometry $T$ such that $T(x) = 0$ (the center of the Poincaré disk $D$), and put $v' = T_*(v)$ and $w' = T_*(w)$. If $C'$ is the diameter of $D$ bisecting the angle between $v'$ and $w'$, then $C = T^{-1}(C')$, the image of $C'$ under the inverse of $T$, is the hyperbolic line through $x$ bisecting the angle between $v$ and $w$.

