Möbius transformation sending line and circle to concentric circles

Question

I want to find some Möbius transformation sending the line $l$ defined by $Re(z) = 5$ and the circle $\vert z\vert = 4$ to concentric circles in the complex plane. I know that Möbius transformations preserve (generalized) circles and that möbius transformations are closed under composition via its group structure.

So the simplest inversion Möbius transformation $T_1(z) = \frac{1}{z}$ turns $l$ into the circle with radius $\frac{1}{10}$ situated at $\frac{1}{10} +0i$. I can then turn it into the unit circle by two composed möbius transformations moving it to the left by $\frac{1}{10}$ and then scale it by $10$. Even if this is correct, how can I make sure that $\vert z\vert = 4$ stays somewhat "invariant" so that encloses the image of my line $l$?

Answer 1

One possible method is to find points $a, b$ which are symmetric with respect to both the given circle $C$ and the given line $l$, and then choose $T$ as a Möbius transformation which sends $a, b$ to $0, \infty$: $$ T(z) = \frac{z-a}{z-b} $$ Möbius transformations preserve symmetry, therefore $0, \infty$ are symmetric with respect to both $T(C)$ and $T(l)$, which are therefore circles centered at zero.

The symmetry of the problem suggest to search for real $a, b \in \Bbb R$. Then symmetry with respect to the line $\operatorname{Re} z = 5$ means $a + b = 10$, and symmetry with respect to the circle $\vert z \vert = 4$ means $ab = 4^2$.

So $a, b$ are the solutions of the quadratic equation $x^2 - 10 x + 16 = 0$, which can easily be determined.

Answer 2

In the desired concentric situation, all lines through the common center will ber perpendicular to both circles. These must correspond to circles/lines that are perpendicular to the given circle and line and intersect in two points. If you bring one of these two points to infinity, you should be done.

So how do we find enough of these common perpendicular circles/lines? One is easy: Take the line through the origin of the circle and perpendicular to the line. Another can be found as circle around any point on the line (so that it at least intersects the line at a right angle) of suitable radius, obtained using Pythagoras from the given circle's radius and the distance.