Four points determine a sphere

To see how this philosophy works, consider four prescribed points $(x_i, y_i, z_i)$, a fifth arbitrary point $(x, y, z)$ and the set of five equations which assert that these five points all lie on the surface of a sphere (setting $x^2+y^2+z^2=r^2$):

$$\begin{eqnarray*} r_1^2\alpha + x_1\beta + y_1\gamma + z_1\delta + \epsilon &=&... ... \\ r^2\alpha + x \beta + y \gamma + z \delta + \epsilon &=&0. \end{eqnarray*}$$

The singularity of the coefficient matrix requires that

$$\begin{eqnarray*} \left\vert \begin{array}{ccccc} r_1^2 & x_1 & y_1 & z_1 & 1 ... ...1 \\ r^2 & x & y & z & 1 \end{array} \right\vert. & = & 0 \end{eqnarray*}$$

Laplace's expansion of this determinant according to its last row yields explicit values for the coefficients $(\alpha, \beta, \gamma, \delta, \epsilon)$. Numerically, relatively arbitrary values can be placed it the last row, and the whole matrix inverted; we want the last row of the inverse, which is insensitive to such choices.