Prove that the locus of a point which moves such that the sum of the square of its distances from the vertices of a triangle is constant is a circle having centre at the centroid of the triangle.

Explanation

Let,`AP^2+AB^2+PC^2=C`

`(x−x_1​)^2+(y−y_1​)^2+(x−x_2​)^2−(y−y_2​)^2+(x−x_3​)^2+(y−y_3​)^2=c`

=`3x^2−(2x_1​+2x_2​+2x_3​)x+x_1^2​+x_2^2​+x_3^2​+3y^2−2(y_1​+y_2​+y_3​)+y_1^2​+y_2^2​+y_3^2​=c`

=`x^2+y^2−2/3​(x_1​+x_2​+x_3​)x−2/3​(y_1​+y_2​+y_3​)c+(x_1^2​+x_2^2​+x_3^2​+y_1^2​+y_2^2​+y_3^2​−c)=0`

So, center of this circle is` ((x_1​+x_2​+x_3)/3​​,(y_1​+y_2​+y_3​​)/3),`

centroid of triangle is a center of that circle.