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

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?

Solution 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 $P$.