let $B_\delta(p):=\{x\in\mathbb{R}^d:||x-p||_2\leq \delta\}$ be a $d$-dimensional closed ball.

Now I do not have one ball, but four: $B_{r_1}(p)$, $B_{r_2}(p)$, $B_{s_1}(q)$ and $B_{s_2}(q)$. Furthermore, the following properties hold:

- $r_1+s_1\geq ||p-q||_2$, thus $B_{r_1}(p)\cap B_{s_1}(q)\neq \emptyset$,
- $r_2+s_2\geq ||p-q||_2$, thus $B_{r_2}(p)\cap B_{s_2}(q)\neq \emptyset$.

Now I want to get an expression for

\[ \sup_{x\in B_{r_1}(p)\cap B_{s_1}(q)}\quad \sup_{y\in B_{r_2}(p)\cap B_{s_2}(q)} \quad ||x-y||_2. \]

I have been racking my head over it, but I do not find an elegant solution. If nothing else helps, I could only solve the problem for $d=2$ and represent the hull of the balls in parametric form and do some minimization there. But I would rather like a more general approach.

Here a small drawing of one of the scenarios that might happen:

The distance of interest here is between the upper intersection of the two red balls ($B_{r_1}(p)$ and $B_{s_1}(p)$) and the lower intersection of the two black balls. Roughly measured, this equals 3.

Any help would be welcome. Thanks in advance.