circumcenter of the $n$-simplex

Question

Given $m$ points $v_i\in\mathbb{R}^n$, $m<n$. How to find the circumcenter of the simplex formed by the points?

Answer 1

In Miroslav Fiedler's lovely book Matrices and Graphs in Geometry, he shows how one can use the Cayley-Menger matrix,

$$\mathbf M=\begin{pmatrix} 0&1&1&\cdots&1\\ 1&0&d_{1,2}^2&\cdots&d_{1,n+1}^2\\ 1&d_{2,1}^2&\ddots&&\vdots\\ \vdots&\vdots&&\ddots&d_{n,n+1}^2\\ 1&d_{n+1,1}^2&\cdots&d_{n+1,n}^2&0\end{pmatrix}$$

to determine the circumsphere of an $n$-simplex determined by $n+1$ $n$-dimensional points. Here, $d_{j,k}$ signifies the distance between vertices $v_j$ and $v_k$. Using the formulae from this page, and letting $\mathbf Q=-2\mathbf M^{-1}$, the circumcenter is given by

$$\left(\frac{q_{1,2}}{q_{1,2}+\cdots+q_{1,n+2}}v_1,\cdots,\frac{q_{1,n+2}}{q_{1,2}+\cdots+q_{1,n+2}}v_{n+1}\right)$$

and the circumradius is given by $\dfrac{\sqrt{q_{11}}}{2}$. The book also discusses how to use the Cayley-Menger matrix to get the insphere. (I wrote up a Mathematica implementation of Fiedler's formulae here.)

Answer 2

the center of the circum-sphere of a simplex can be trivially calculated as the solution of a sytem of linear of equations: calculate any spanning tree of the corners and take the intersection of the hyperplanes that contain the midpoints of the tree edges and are orthogonal to them.