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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.

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2 Answers 2

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

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  • $\begingroup$ What if there were 6 points? $\endgroup$
    – user541686
    Commented Oct 18, 2014 at 20:07
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    $\begingroup$ @Mehrdad Then you can do it with two fives of them $\endgroup$
    – kinokijuf
    Commented Oct 18, 2014 at 20:41
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    $\begingroup$ Why does this work? $\endgroup$
    – user66698
    Commented Oct 19, 2014 at 5:34
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    $\begingroup$ Note that this detects coplanarity as well; depending on your application, you may need to do a separate check that the five points are not coplanar. $\endgroup$
    – Mark Dickinson
    Commented Oct 19, 2014 at 9:22
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    $\begingroup$ @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. $\endgroup$
    – Mark Dickinson
    Commented Oct 19, 2014 at 9:29
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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.

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