How to show that five points in ℝ³ are cospherical?

There are many conditions equivalent to the cocircularity of four points on a plane, however i could not find any such lists for the three-dimensional analog. When do five points in three-dimensional space lie on one sphere?

Elementary conditions preferred.

When $\begin{vmatrix} x_1^2+y_1^2+z_1^2&x_1&y_1&z_1&1\\ x_2^2+y_2^2+z_2^2&x_2&y_2&z_2&1\\ x_3^2+y_3^2+z_3^2&x_3&y_3&z_3&1\\ x_4^2+y_4^2+z_4^2&x_4&y_4&z_4&1\\ x_5^2+y_5^2+z_5^2&x_5&y_5&z_5&1\\ \end{vmatrix}=0$.
• @Emrakul: if the determinant is zero, there's a nonzero vector $(a, b, c, d, e)$ in the (right) nullspace of the given matrix, so every point $(x, y, z)$ in our set of five points lies on the surface $a(x^2 + y^2 + z^2) + bx + cy + dz + e = 0$. If $a$ is nonzero, it's easy to rearrange this into the equation of a sphere with centre $(-b/2a, -c/2a, -d/2a)$. (If $a$ is zero then the five points are coplanar rather than cospherical.) The other direction is similar: if the five points lie on a sphere then you can find a nonzero vector in the nullspace of the given matrix. – Mark Dickinson Oct 19 '14 at 9:29