A set of points is contained in a sphere $S$. When is $S$ also the circumsphere?

Question

Given points $p_1,\ldots,p_n\in\Bbb R^d$ so that all of them are contained in a common sphere $S\subset\Bbb R^d$ (by which I mean the usual $(d-1)$-dimensional sub-manifold of $\Bbb R^d$). Note that $S$ is not necessarily the circumsphere of the set $\{p_1,\ldots,p_n\}$, by which I mean the smallest sphere that bounds a ball that contains all $p_1,\ldots,p_n$:

But I wonder, when do $S$ and the circumsphere agree? More specifically, I wondered whether the following seemingly elementary question has a positive answer, and a simple proof:

Question: If the convex hull $\mathrm{conv}\{p_1,\ldots,p_n\}$ contains the center of $S$, is then $S$ also the circumsphere?

Answer 1

Yes.

WLOG, let $S$ be of radius $1$ and centered at the origin. Let $D$ be the closed ball with boundary $S$, let $C$ be the circumsphere with center $c$, and let $P$ be the set of points $p_i$. $C$ is at most as large as $S$, so $P\subset D+c$. Then for each $i$: $$1\geq|p_i-c|^2=|p_i|^2 +|c|^2-2p_i\cdot c=1+|c|^2-2p\cdot c$$ Rearranging, $$p_i\cdot c\geq \frac{|c|^2}{2}$$ Now if the origin is contained within the convex hull of $P$, then there is some linear combination $$\sum_i\lambda_ip_i=0\quad\quad\quad \sum_i\lambda_i=1$$ However, we must also have $$0=c\cdot \left(\sum_i\lambda_ip_i\right)\geq \frac{|c|^2}{2}$$ So $C=S$ if the center of $S$ is contained in the convex hull of $P$.

Answer 2

For each $i\in\{1,...,n\}$ the map $\Bbb R^d\ni x\mapsto \|x-p_i\|^2$ is strictly convex, and so is their maximum

$$x\;\mapsto\; \max_i \|x-p_i\|^2 =: [r(x)]^2.$$

Note that $r(x)$ is exactly the radius of the smallest sphere with center $x$ that bounds a ball that contains all $p_i$. From strict convexity follows the existence of a unique minimizer $x^*$ of $[r(x)]^2$, which is the center of the circumsphere.

Now, let $s$ be the center of $S$, and suppose that $s\in\mathrm{conv}\{p_1,...,p_n\}$. Assume further that $v:=x^*-s\not=0$. Then, since $[r(x)]^2$ is strictly convex, $[r(s+\epsilon v)]^2<[r(s)]^2$ for all sufficiently small $\epsilon >0$. But since $s$ is in the convex hull, there is a $p_i$ with $\langle p_i-s,v\rangle \le 0$, and thus

$$[r(s+\epsilon v)]^2 \ge \|s+\epsilon v - p_i\|^2=\underbrace{\|s - p_i\|^2}_{=\,[r(s)]^2}+\|\epsilon v\|^2+2\epsilon \underbrace{\langle s - p_i, v\rangle}_{\ge\, 0} \ge [r(s)]^2.$$

This is a contradiction. Thus $v=0\implies s=x^*$ and $S$ is the circumsphere.