Given two disks $D_1\subsetneq D_2$ such that $\partial D_1\cap \partial D_2 \neq \emptyset$, then the intersection consists of exactly one point. I'm reading Rudin's Real and Complex Analysis, and I'm stuck at the proof of theorem 10.37, which essentially shows that the index of a curve increases by $1$ as we move "from right to left". The problem is as follows:
Let $r>0$ and $a,\zeta \in \mathbb{C}$, with $|a-\zeta|=r$. Consider a set $E\subset \overline{D(a,r)}$ with $\zeta \not \in E$. He goes on to state that $E\subset D(2a-\zeta,2r)$. 
Now, its easy to show that $E\subset \overline{D(2a-\zeta,2r)}$, but I can't seem to figure out how to prove that it can't lie in the boundary of said disk. I was able to show that $D(a,r)\subset D(2a-\zeta,2r)$, and that if a point $z\in \overline{D(a,r)}$ also belongs to $\partial D(2a-\zeta,2r)$, then it must lie in the intersection of the boundaries, which intuitively consists of a single point ($\zeta$) since one disk is a proper subset of the other one. I'm looking for a short proof of the last assertion. I suppose one should be able to do it with the equation of two circles in the plane, but that seems quite tedious. 
Any thoughts?
Thanks in advance!
 A: $D(2a-\zeta,2r)$ contains the entire $\overline {D(a,r)}$, except $\zeta$: this is because $2a-\zeta$ is the point on the boundary of $D(a,r)$ opposite to $\zeta$. If you're still unconvinced, just draw a picture. Or consider without loss of generality the case when $a=0$, $r=\zeta=1$.
This is just a matter of elementary plane geometry, no need to involve complex numbers here.
A: You need the fact that equality holds in the complex triangle inequality
$| z+w| \le |z| + |w|$ exactly if $z=0$ or $w=0$ or $\frac zw$ is a positive
real number ($z$ and $w$ have the same argument).
For $z \in E$, $|z-a| \le r$ and therefore
$$
  | z - (2a-\zeta)| = | (z-a) + (\zeta - a)| \le | z-a | +  |\zeta - a|
 \le r + r = 2r \, .
$$
Now assume that equality holds in this inequality chain. Then $| z-a | =|\zeta - a| = r$ and
$$
\frac{z-a}{\zeta - a} = \lambda > 0
$$
It follows that $\lambda = 1$ which implies
$z = \zeta$, in contradiction to the assumption that $\zeta \not \in E$.
So equality cannot hold, i.e. $| z - (2a-\zeta)| < 2r$ which is
what you wanted to show.
