Find the locus of points, the sum of the squares of the distances from which to the vertices of an equilateral triangle

Find the locus of points, the sum of the squares of the distances from which to the vertices of an equilateral triangle is equal to the square of the perimeter of this triangle.

Consider an equilateral triangle, each side of which is equal to a. In order to find the locus of points, the sum of the squares of the distances from which to the vertices of an equilateral triangle is equal to the square of the perimeter of this triangle (that is, 3 * a), we apply the so-called coordinate method.
Suppose that one vertex of this equilateral triangle is at the origin (point O (0; 0)), and the other vertex (A) has coordinates (a; 0). In the case when a = 10.
Let’s denote by B the third vertex of an equilateral triangle and find its coordinates (xb; yb). By construction, the abscissa xb of the vertex B is equal to the arithmetic mean of the abscissa of the points O (0; 0) and A (a; 0), that is, xb = (0 + a) / 2 = a / 2. It is easy to see that the ordinate уb of the vertex B is equal to the height of the triangle ОАВ, that is, уb = (√ (3) * a) / 2.
Let us assume that M (x; y) is some point of the required locus of points. Then, according to the terms of the assignment, AM² + BM² + OM² = (3 * a) ² = 9 * a². Using the formula for calculating the distance between two points, we have: AM = √ ((x – a) ² + (y – 0) ²); BM = √ ((x – a / 2) ² + (y – (√ (3) * a) / 2) ²) and ОM = √ ((x – 0) ² + (y – 0) ²). So, (x – a) ² + (y – 0) ² + (x – a / 2) ² + (y – (√ (3) * a) / 2) ² + (x – 0) ² + (y – 0) ² = 9 * a².
Let’s open the brackets: x² – 2 * x * a + a² + y² + x² – 2 * x * (a / 2) + (a / 2) ² + y² – 2 * y * (√ (3) * a) / 2 + ((√ (3) * a) / 2) ² + x² + y² = 9 * a². Let us transform the resulting equation. Then, we get: (1 + 1 + 1) * x² + (-2 – 1) * a * x + (1 + 1 + 1) * y² – √ (3) * a * y = 9 * a² – a² – a² / 4 – (3 * a²) / 4 or 3 * x² – 3 * a * x + 3 * y² – √ (3) * a * y = 7 * a².
Let us bring the resulting equation to the form: (x – a / 2) ² + (y – (√3 * a) / 6) ² = ((2√6 * a) / 3) ². That is, the required locus of points is a circle centered at the point C ((a / 2); (√3 * a) / 6) of radius R = (2√6 * a) / 3. It can be proved that C is the center of the circle is also the center of this triangle.