$26 \cdot 6$ Given an equilateral triangle with side length $s$, $P$ is a point in the plane of the triangle, and the sum of the squares of the distances from $P$ to the three vertices of the triangle is a constant $a$. The locus of all points $P$ that satisfy this condition is
(A) If $a>s^{2}$, it is a circle.
(B) If $a=2 s^{2}$, it contains only three points; if $a>2 s^{2}$, it is a circle.
(C) It is a circle with a positive radius only when $s^{2}<a<2 s^{2}$.
(D) For any value of $a$, it contains only a finite number of points.
(E) None of these.
(27th American High School Mathematics Examination, 1976)
[Solution] Let the coordinates of point $P$ in the coordinate system be $(x, y)$, and the vertices of the equilateral triangle be $(0,0), (s, 0)$, and $\left(\frac{s}{2}, \frac{\sqrt{3} s}{2}\right)$. Then the trajectory of point $P$ satisfies
$$
a=x^{2}+y^{2}+(x-s)^{2}+y^{2}+\left(x-\frac{s}{2}\right)^{2}+\left(y-\frac{\sqrt{3} s}{2}\right)^{2} \text {, }
$$

which is $\square$
or
$$
\begin{array}{l}
a=\left(3 x^{2}-3 s x\right)+\left(3 y^{2}-s \sqrt{3 y}\right)+2 s^{2}, \\
\frac{a-2 s^{2}}{3}=\left(x-\frac{s}{2}\right)^{2}+\left(y-\frac{\sqrt{3} s}{6}\right)^{2}-\frac{s^{2}}{3} .
\end{array}
$$

which is $\frac{a-s^{2}}{3}=\left(x-\frac{s}{2}\right)^{2}+\left(y-\frac{\sqrt{3} s}{6}\right)^{2}$.
Thus, if $a > s^{2}$, the trajectory is a circle.
Therefore, the answer is $(A)$.