Existence of unique circle passing through interior points of unit disk meeting the boundary orthogonally I am a self-studies and this is a hw problem from a complex analysis scourse I've been doing.
The problem set pertains to the topic Automorphism Groups and has a high concentration of fractional linear transformations. So I would be appreciative of any help, but especially if those concepts are applicable.


Show that for any points $a, b\in D$, the unit disk, there is a unique circle $C$ passing through $a$ and $b$ and meeting $\partial D$ orthogonally. (Suggestion: Prove first that the only circles through $0$ and perpendicular to $\partial D$ are diameters of D.)


As far as the hint goes, it makes sense that radii are perpendicular to tangents to the circle, so if the radii lie on a line, it is a diameter. But this doesn't seem like much of a proof.
Also as far as the general case, I can see how a circle through $a$ and $b$ and centered outside of $D$ can have orthogonal intersections with the $\partial D$ and how the respective right triangles formed by $0$ and the two tangents to the circle from the center of $C$ work.
But I would appreciate help proving the existence and uniqueness. Especially if there is a way using fractional linear transformations.
Thanks
 A: different problems may be easiest in different models. This one is best in the upper half plane. You need to know that any Mobius transformation preserves angles (including right angles) and maps any line or circle into either a line or circle. 
To take the unit disc to the upper half plane, use
$$  f(z) = \frac{z + i}{iz+1}. $$
To take the upper half plane to the disc, use
$$  g(z) = \frac{iz + 1}{z+i}. $$
Which, now that I look at them together, are just reciprocals. 
In the upper half plane, given two distinct points, there are just two possibilities. If two points are on a vertical ray, because both have the same real part, that is the line through them. If the two points have different real parts, draw the ordinary line segment between them, draw the perpendicular bisector of that segment, it meets the real axis at a point, call it $P.$ The semicircle with center at $P$ that passes through both original points is what you want.  
In both cases, the curve described is mapped to the unit disc to either a diameter or a circular arc that is orthogonal to the boundary. It is worth the exercise to find out what happens to $z = A + i e^t$ under the mapping $  g(z) = \frac{iz + 1}{z+i}. $ Here $A$ is real constant, $t$ the real variable.
