# 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 $\begin{vmatrix} x_1^2+y_1^2+z_1^2&x_1&y_1&z_1&1\\ x_2^2+y_2^2+z_2^2&x_2&y_2&z_2&1\\ x_3^2+y_3^2+z_3^2&x_3&y_3&z_3&1\\ x_4^2+y_4^2+z_4^2&x_4&y_4&z_4&1\\ x_5^2+y_5^2+z_5^2&x_5&y_5&z_5&1\\ \end{vmatrix}=0$.

• What if there were 6 points? Oct 18 '14 at 20:07
• @Mehrdad Then you can do it with two fives of them Oct 18 '14 at 20:41
• @kinokijuf: fair enough, but I meant is there a more elegant way than something like that. Oct 18 '14 at 20:46
• Why does this work?
– user66698
Oct 19 '14 at 5:34
• @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. Oct 19 '14 at 9:29

Up to a circular inversion, to test five points to be cospherical is the same as testing four points to be coplanar. In terms of mutual distances, this task can be achieved by using the Cayley-Menger determinant.