Given a set of points $X \subseteq \mathbb{R}^d$, where $2\leq|X|\leq d+1$, how would one go about computing the center and radius of the $(d-1)$ sphere with the smallest radius where these points in $X$ lie?
For 2 points, $\vec{x}_0$ and $\vec{x}_1$, we find the midpoint $\frac{\vec{x}_0+\vec{x}_1}{2}$, which is the center point, and the radius $r = \frac{\|\vec{x}_0 - \vec{x}_1\|_2}{2}$
So, for any $3\leq k+1\leq d+1$ points, would the center of this hypersphere always be the centroid: $$\overline{X} = \frac{1}{k+1}\sum_{i=0}^{k} \vec{x}_i$$ with a radius being the Euclidean distance from the centroid to any point in $X$? How would I prove that all of these points are equidistant from the centroid? Or is am I thinking about this wrong? I think I'm wrong...
Let's say we have three points $(1,0),(-1,0),(0,1)$, and the center would be $(0,\frac{1}{3})$ with distances $\sqrt{1 + \frac{1}{9}} = \frac{\sqrt{5}}{3}$, $\sqrt{1 + \frac{1}{9}} = \frac{\sqrt{5}}{3}$, and $\sqrt{0 + \frac{4}{9}} \neq \frac{\sqrt{5}}{3}$, so I'm no longer sure about my reasoning.
Given three points A,B,C in the plane, not collinear, the circle you want has its center where the perpendicular bisectors of the three triangle edges meet. Call that point P.
Next, take those three points ABC, circle center P, but now add a fourth point D off the original plane. Every sphere with the three original points on boundary contains the circle we made. It follows that the center of the sphere we want lies along a line through P but perpendicular to the plane containing ABC. Then take three different points, maybe ABD, in that plane find circle center Q, draw a line through Q perpendicular to ABD. These two lines will meet in the center of the sphere we want.
Sphere-sphere intersection is not a surface
For dimension four, keep going.