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Jan 20, 2022 at 22:19 comment added Saúl RM The algorithm you mention only uses that the distance from your point to the rest of the points is equal, so it will obtain infinitely many solutions if $N<M+1$. This is the case when you are given three non-colinear points in $\mathbb{R}^3$, then there will be a whole line of points at the same distance of the three points. But only one of them is closest to the three points: in this case it can be obtained by intersecting the line of points equidistant to the three given points with the plane containing those points (this is what my answer generalizes).
Jan 20, 2022 at 11:05 comment added Menachem More concretely, I think the algorithm referenced above, that involves solving a system of linear equations, will work just as well for the $N < M+1$ case, though the matrix is no longer square. Assuming the system has a solution, then the point $p$ obtained is the center of the (translated) sphere, and hence the equidistant point you seek.
Jan 20, 2022 at 11:02 comment added Menachem If three distinct points lie on a line, then no point is equidistant to them. If they don't lie on a line, then a unique circle passes through them; its center is the unique equidistant point. The same is true in general. We need to check that the $N$ points are in "general position", i.e., that they don't lie on a hyperplane of dimension lower than $N-1$. Three points shouldn't lie on a line, four points shouldn't lie in a plane etc. If the $N$ points are in general position, then a unique hypersphere passes through them; its center is the equidistant point.
Jan 18, 2022 at 18:56 comment added Erik I like this solution very much for cases where N >= (M+1)! But not sure if I could expand it to cases where N < (M+1)... where there are more than one equidistant points, but one of them is closest?
Jan 18, 2022 at 9:14 history answered Menachem CC BY-SA 4.0