Given the unit circle(without boundary), and three points in the unit disk, $0,a,z$, where $a$ lies on the real axis, $z$ is some arbitrary complex number. I want to find the parametrization for the arc that passes $a$ and $z$ which intersects orthogonally with the unit disk.
My way of doing it is just suppose the equation of the arc is $(x-c)^2 +(y-d)^2= 1$, then solves the intersection point of it with the unit disk in terms of $c$ and $d$, then utilizes the orthogonal condition to solve $c$ and $d$. The formula is very nasty even for the intersection points, so I am not sure that this will work right. Are there simpler ways?
The trick is that the Möbius transformation $$ T(w) = \frac{w-a}{1-\bar{a}w} $$ maps the unit circle to itself, $T(a)=0$. (Also, since $a$ is real, $\bar{a}=a$.) Möbius transformations preserve orthogonality, and we know that circular arcs through $0$ that are perpendicular to the unit circle are just radial.
So, we want a parametrisation of the radial line through $T(z)$. We then undo the transformation to get the original arc.
So, the radial line through the image of $z$ is $tT(z)$. So we have $$ T^{-1}(tT(z)) = \frac{tT(z)+a}{1+atT(z)} = \frac{t(z-a)+a(1-az)}{(1-az)+at(z-a)}. $$ This clearly passes through $a$ when $t=0$ and $z$ when $t=1$.