Circumcenter of Tetrahedron (in 4D)

Question

I am trying to calculate the circumcenter of a tetrahedron in 4 dimensional space. I was hoping for some concrete mathematical formula which can make this calculation more accurate.

Answer 1

This goes well beyond your question, but this paper was posted to the arXiv just last week:

Jean-Daniel Boissonnat, Ramsay Dyer, Arijit Ghosh. "A probabilistic approach to reducing the algebraic complexity of computing Delaunay triangulations." (arXiv abstract).

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Computing Delaunay triangulations in $\mathbb{R}^d$ involves evaluating the so-called in_sphere predicate that determines if a point $x$ lies inside, on or outside the sphere circumscribing $d+1$ points $p_0,\ldots,p_d$. This predicate reduces to evaluating the sign of a multivariate polynomial of degree $d+2$ in the coordinates of the points $x,p_0,\ldots,p_d$. [...]

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Searching on the phrases in_sphere or in-sphere or insphere will access the substantial literature on the topic.

Answer 2

Let the points be $(a1,a2,a3,a4),(b1,b2,b3,b4),(c1,c2,c3,c4),(d1,d2,d3,d4)$.
The circumcentre is in the hyperplane, so it is a convex combination of the points $$CC = wA+xB+yC+(1-w-x-y)D$$ The distances to the points are the same. $$|(1-w)AD-xBD-yCD|^2\\=|-wAD+(1-x)BD-yCD|^2\\=|-wAD-xBD+(1-y)CD|^2\\=|-wAD-xBD-yCD|^2$$
Subtract the last expression from the first three, to get three linear equations in $w,x,y$:
$$\left[\begin{array}{ccc}2AD.AD&2AD.BD&2AD.CD\\2BD.AD&2BD.BD&2BD.CD\\2CD.AD&2Cd.BD&2CD.CD\end{array}\right]\left[\begin{array}{c}w\\x\\y\end{array}\right]=\left[\begin{array}{c}AD.AD\\BD.BD\\CD.CD\end{array}\right]$$
For example, $2AD.BD=2(d-a).(d-b)$. Invert the 3x3 matrix, to find $w,x$ and $y$.