The point P moves such that the sum of the square of the distances from the vertices of a triangle ABC is constant, then thelocus of P is a circle with centre is ................... of triangle ABC

Question

This is a bit of the chapter  LOCUS  of first year intermediate

Answer 1

Let the vertices of the triangle be A(a,b),B(c,d), C(e,f). Let P(x,y) be a point on the locus.

The distance of P from each vertex is given by sqrt((x-a)^2+(y-b)^2), sqrt((x-c)^2+(y-d)^), sqrt((x-e)^2+(y-f)^2), and the sum of the squares of these distances is K=3(x^2+y^2)+(a^2+b^2+c^2+d^2+e^2+f^2)-2x(a+c+e)-2y(b+d+f).

So K-(a^2+b^2+c^2+d^2+e^2+f^2))/3=x^2-2x(a+c+e)/3+y^2-2y(b+d+f)/3=

(x-(a+c+e)/3)^2+(y-(b+d+f)/3)^2-(a+c+e)^2/9-(b+d+f)^2/9.

Thus, K-(a^2+b^2+c^2+d^2+e^2+f^2))/3+(a+c+e)^2/9+(b+d+f)^2/9=(x-(a+c+e)/3)^2+(y-(b+d+f)/3)^2.

The centre of this circle, circumcentre of the triangle, is ((a+c+e)/3,(b+d+f)/3). Note that the coords are the average of the coords of the vertices.

I think this is the circumcircle of the triangle, or the circumscribed circle.