Möbius transformation sending line and circle to concentric circles

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 $l$?

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.