Number of disjoint circles in half plane minus a disk that touch both boundary components 
Let $\Omega \subset \mathbb{C}$ be right half-plane, with the disc $D$ removed, where $D$ is the disk of radius $r=3$ centered at $z_0=5$. What is the maximum number of disjoint open disks in $\Omega$ whose boundary touch both boundary components of $\Omega$?

This is a question I had on my qualifying exam years ago; I couldn't solve it then, and today I stumbled upon it again and it nags me that I can't find a solution. Or at least, one without using any table or any map that I couldn't come up with on my own during an exam.
What I did (then and now) is to transform conformally the right half plane into the unit circle, and then see where the disk $D$ is mapped. Then my idea was to count the number of circles in this scenario. If the circles are not those coming from vertical lines in the right half plane, then they will come from circles in the original domain $\Omega$ (a simple test for a circle not to be a "bad" circle, is not to be centered on the real line, for example).
That is, 
$$
f(z)=\frac{z-1}{z+1}
$$
maps $\Omega$ to the unit disk with the disk $\tilde{D}$ removed, where $\tilde{D}$ is the disk centered at $w_0=\frac{11}{15}$ of radius $\frac{2}{15}$ (this follows from some elementary but tedious computations which I won't include in the question).
Now, I don't know how many disjoint circles touching both boundary components of $f(\Omega)$ I can fit in it, nor do I know how to guarantee that the answer is invariant under $f^{-1}$.
 A: As already said in the comments, it is useful to map $\Omega$
with Möbius transformation $T$ to an annulus
between two concentric circles, and we can assume that annulus is centered  around $w=0$.
Some geometric observations help to find such a transformation $T$.
Let $a$ and $b$ be the pre-images of $\infty$ and $0$, respectively.


*

*From the symmetry of $\Omega$ and the annulus with respect to the 
real axis we can conclude that both $a$ and $b$ are real.

*$a$ and $b$ must be symmetric with respect to the imaginary axis, i.e.
$b = -\overline a$. For real $a, b$, this simplifies to $b = -a$.

*$a$ and $b$ must be symmetric with respect to the circle of
radius $r=3$ centered at $z_0 = 5$, i.e. $(a - z_0)\overline{(b-z_0)} = r^2$. In our case, this simplifies to $(a-5)(b-5) = 3^2$.


It follows that $9 = (a-5)(-a-5) = -a^2 + 25$ and therefore $a = \pm 4$,
so
$$
 T(z) = \frac{z+4}{z-4}
$$
is a candidate. Now verify that $T$ indeed maps $\Omega$ to an annulus
with center $w=0$, inner radius $R_1= 1$ and outer radius $R_2 = 3$.
Disks touching both boundary components of $\Omega$ are mapped by $T$ to
disks touching both inner and outer circle of the annulus.
These image disks must have radius one and a center on $|w| = 2$. It is now
easy to see that there can be at most six of them,
compare Wikipedia: Steiner chain 
(link provided by @ccorn in a comment)
or Wolfram World: Steiner chain.
A: Just adding a graphical figure to Martin R's answer.
The transformation used in the figure differs slightly from Martin's in that
$$f(z)=4+\frac{16}{z-4}=\frac{4z}{z-4}=2\left(1+T(z)\right)$$
but both are equally usable. The important point is that the inversion circle has its center at $c=4$. Well, $c=-4$ would have worked as well, both possibilities follow from requiring constant annulus width:
$$f(0)-f(z_0-r) = f(z_0+r)-f(\infty)$$

The solid black circles are $\partial D$
and an exemplary set of solution circles.
The dash-dotted circle represents the inversion circle associated with
$\bar{f}$.
The gray circles are the boundaries of $f(\Omega)$.
The dashed circles are the images of the solution circles under $f$,
these form a Steiner chain with respect to the gray circles.
Two of the dashed circles disappear under solid circles,
those are just exchanged by $f$.
Note that the contents of the annulus could be rotated and still yield
valid solution circles under the inverse transform, with one exception: No dashed circle shall touch the inversion center $c$, otherwise it maps back
to a (vertical) line which does not touch the $y$-axis
(in $\mathbb{C}$, that is).
