A problem to show a certain sum is invariant in some Euclidean geometric configurations. $\textbf{Problem.}$ Suppose circles of radius $r$ and radius $s$ are externally tangent at the point $1/2$ and internally tangent to the unit circle. There are infinitely many such configurations, one of which is illustrated in the diagram below. Prove that $\displaystyle\frac{1}{r}+\frac{1}{s}=\frac{16}{3}$ for every such configuration.

The above problem appeared in a complex analysis exam in some graduate school. I tried number of things, but nothing worked yet and I have little clue about what I should do to solve this problem.
One of the thing I tried is just setting the given configuration into the following equations by setting the centers of the circles as $p,q$ to obtain $|p-\frac{1}{2}|=r,|q-\frac{1}{2}|=s,|p|+r=|q|+s=1,r+s=|p-q|$ and to try some random computations but it didn't work.
I also tried to use some Mobius transform but it didn't work. Please let me know how to solve this or if this problem is related to some known stuff.
 A: $(1)$ Formulae for $r$ and $s$:
 See the diagram below. Since $|1/2-(1-r)z_0|=r$, we have
$$
\frac{1}{4}-(1-r)\operatorname{Re}\,z_0+(1-r)^2=r^2,$$
$$
r=\frac{5-4\cdot\operatorname{Re}\,z_0}{8-4\cdot\operatorname{Re}\,z_0}.$$
Simirarly we have
$$
s=\frac{5-4\cdot\operatorname{Re}\,z_1}{8-4\cdot\operatorname{Re}\,z_1}.$$

$(2)$ We consider a general Mobius transformation $w=f(z)$ which maps the unit disk onto itself and fixes the point $\frac{1}{2}$. 
To find $f$ we consider $\varphi, \phi, g$ as follows: 
\begin{align}
\zeta&=\varphi (z)=\frac{2z-1}{2-z},\quad \varphi (1/2)=0, \; \varphi (1)=1,\;\varphi (-1)=-1\\
\xi&=\phi(\zeta)=e^{i\theta }\zeta,\;(\text{rotation})\\
w&=g(\xi)=\frac{2\xi+1}{\xi+2},\quad g(0)=1/2,\; g(e^{i\theta })=z_0,\;g(-e^{i\theta })=z_1.
\end{align}
Their composition$$
w=f(z)=g\circ \phi\circ \varphi (z)=\frac{(4e^{i\theta }-1)z+2(1-e^{i\theta })}{(4-e^{i\theta })+2(e^{i\theta }-1)z}$$
is a Mobius transformation we are looking for.
See the diagram below.

Note that ''every such configuration'' is the image of disks $|z-3/4|=1/4$ and $|z+1/4|=3/4$ by $w=f(z)$ above. We see that $$
z_0=f(1)=\frac{1+2e^{i\theta }}{2+e^{i\theta} }, \quad \operatorname{Re}\, z_0=\frac{4+5\cos \theta }{5+4\cos\theta }$$ and $$
z_1=f(-1)=\frac{1-2e^{i\theta }}{2-e^{i\theta }}, \quad\operatorname{Re}\, z_1=\frac{4-5\cos \theta }{5-4\cos\theta }.
$$
Therefore we have $$
r=\frac{3}{8+4\cos\theta }, \quad s=\frac{3}{8-4\cos\theta },$$
which yields
$$
\frac{1}{r}+\frac{1}{s}=\frac{16}{3}$$
 for every $\theta $. The proof is complete.
