Denote the unit circle with $\mathbb{D}$. Let $\zeta_1, \zeta_2 \in \partial\mathbb{D},\, \zeta_1\neq\zeta_2$ and $C=K_r(a)$ (circle radius $r$ around the point $a$), which intersects $\partial\mathbb{D}$ in $\zeta_1$ and $\zeta_2$ orthogonally. $M$ is the set of all Möbius transformations that map $\mathbb{D}$ conformally to the upper half plane ($\mathbb{H}=\{z\in\mathbb{C} \mid \operatorname{Im}(z)>0 \}$). The reflection point $p^*$of $p$ on the „hyperbolic line“ $C\cap \mathbb{D}$ is defined by $p^*=T^{-1}(\overline{T(p)})$ for $T\in M$.
Prove that $p^*$ is well defined, i.e. independent of the the chosen Möbius transformation.
My first idea was to somehow map the whole system onto the upper halfplane and then have $C\cap\mathbb{D}$ to a line. There the reflection is easier to think of an then transform it back, but I have no clue how to work on this problem more specifically. I would be grateful for any help.
If $T, S \in M$ are two Möbius transformations mapping the unit disk $\Bbb D$ onto the upper halfplane $\Bbb H$ then $\phi = S \circ T^{-1}$ is an automorphism of $\Bbb H$, and these satisfy $$ \overline{\phi(w)} = \phi(\overline w) \, . $$ (This follows from the fact that $\phi(w) = \frac{aw+b}{cw+d}$ with real numbers $a, b, c, d$. It is also a consequence of the Schwarz reflection principle.)
Now assume that $T(p^*) = \overline {T(p)}$. Then also $$ S(p^*) = \phi(T(p^*)) = \phi(\overline {T(p)}) = \overline {\phi(T(p))} = \overline {S(p)} $$ and that proves that the definition of reflection points is independent of the chosen Möbius transformation in $M$.