Prove that the locus of the point that moves such that the sum of the squares of its distances from the three vertices of a triangle is constant is a circle.

Explanation

Let vertices of triangle are (0,0), (a,0) and (b,c).
The variable point (x,y) is such that
`(x-o)^(2)+(y-0)^(2)+(x-a)^(2)+(y-0)^(2)+(x-b)^(2)+(y-c)^(2)=lambda`(constant)
`:. 3x^(2)_3y^(2)-2(a+b)x-2cy+a^(2)+b^(2)+c^(2)-lambda=0`, which is equation of circle.