= Solution
Use <vectors> in three-dimensional <Euclidean space>, with the usual <dot product> and <cross product>. The following explicit assignments compute the <circumcenter of a tetrahedron>:
$$
\begin{aligned}
u&\leftarrow B-A,&v&\leftarrow C-A,&w&\leftarrow D-A,\\
c_u&\leftarrow v\times w,&c_v&\leftarrow w\times u,&c_w&\leftarrow u\times v,\\
d&\leftarrow u\cdot c_u,&a&\leftarrow u\cdot u,&b&\leftarrow v\cdot v,&c&\leftarrow w\cdot w,\\
x&\leftarrow\frac{a c_u+b c_v+c c_w}{2d},&O&\leftarrow A+x.
\end{aligned}
$$
General position means $d\ne0$. The reciprocal cross-product identities give $x\cdot u=|u|^2/2$, $x\cdot v=|v|^2/2$ and $x\cdot w=|w|^2/2$. Consequently
$$
|O-B|^2-|O-A|^2=|x-u|^2-|x|^2=|u|^2-2x\cdot u=0,
$$
and likewise for $C,D$. Thus \b[$O$ is the unique point equidistant from the four vertices]. It can lie outside the <tetrahedron>; no inside constraint should be imposed.
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