Circle Center From Three Points (Expanded)

The circumcentre $X$ of the triangle ABC, whose vertices have position vectors $\mathbf{a},\mathbf{b},\mathbf{c}$ respectively, has the following properties:

• it is equidistant from $A$ and $B$ , and hence lies on the plane passing through the midpoint of $AB$ which is normal to $AB$ . Thus it lies on the plane with vector equation $\mathbf{r} \cdot (\mathbf{b}-\mathbf{a}) \,=\, \tfrac12(|\mathbf{b}|^2 - |\mathbf{a}|^2)$ .
• it is equidistant from $A$ and $C$ , and hence lies on the plane passing through the midpoint of $AC$ which is normal to $AC$ . Thus it lies on the plane with vector equation $\mathbf{r} \cdot (\mathbf{c}-\mathbf{a}) \,=\, \tfrac12(|\mathbf{c}|^2 - |\mathbf{a}|^2)$ .

With the points $A\;(-1,7,5)$ , $B\; (13,9,2)$ , $C\;(4,-8,1)$ , the intersection of these two planes is the line $\ell$ with equation $x \; = \; \frac{2697 + 106 u}{440} \hspace{1cm} y \; = \; \frac{811 - 82 u}{440} \hspace{1cm} z \; = \; u$ Moreover $AX$ must lie in the plane $ABC$ , and hence must be perpendicular to the vector $\mathbf{n} \; = \; (\mathbf{b} - \mathbf{a}) \wedge (\mathbf{c} - \mathbf{a}) \; = \; \left(\begin{array}{c} -53 \\ 41\\ -220 \end{array}\right)$ (note that the line $\ell$ is in the direction of the vector $\mathbf{n}$ ). The point on $\ell$ with this final property is found when $u = \frac{22471}{10578}$ , and hence $X$ has coordinates $\left(\frac{351259}{52890}\,,\, \frac{ 1867}{1290}\,,\, \frac{22471}{10578} \right)$ which makes the answer $\boxed{7.1214}$ .