smc

We can consider the locus of points $P$ as the set of points satisfying the equation: $[(x_1-x_2{)}^2+(y_1-y_2{)}^2+(x_1-x_3{)}^2+(y_1-y_3{)}^2+(x_2-x_3{)}^2+(y_2-y_3{)}^2=a]$ where $(x_1,y_1)$, $(x_2,y_2)$, and $(x_3,y_3)$ are the coordinates of the three vertices of the equilateral triangle. If we simplify this equation, we get: $[2(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2)-2(x_1{x}_2+x_1{x}_3+x_2{x}_3+y_1{y}_2+y_1{y}_3+y_2{y}_3)=a]$ Since the vertices of the triangle are fixed, the left side of this equation is also fixed. Thus, the locus of points $P$ is a circle centered at the centroid of the triangle with radius $\sqrt{a}$. We can now determine which of the answer choices is correct: $\text{(A) }\text{is a circle if }a>s^2$ - This is correct, as the radius of the circle will be greater than zero when $a>s^2$. $\text{(B) }\text{contains only three points if }a=2s^2\text{ and is a circle if }a>2s^2$ - This is incorrect, as the locus of points $P$ is always a circle. $\text{(C) }\text{is a circle with positive radius only if }{s}^2<a<2s^2$ - This is incorrect, as the locus of points $P$ is always a circle. $\text{(D) }\text{contains only a finite number of points for any value of }a$ - This is incorrect, as the locus of points $P$ is always a circle. $\text{(E) }\text{is none of these}$ - This is incorrect, as the locus of points $P$ is always a circle. Therefore, the correct answer is $\boxed{\text{is a circle if }a>s^2}$.