Question

The locus of a point which moves such that the sum of the squares of its distances from three vertices of a triangle ABC is constant is a circle
whose centre is at the

• A. centroid of triangle ABC
• B. Circumcentre of triangle ABC
• C. Orthocentre of triangle ABC
• D. incentre of triangle ABC

Answer 1

The correct option is A centroid of triangle ABC

$A{P}^2+A{B}^2+P{C}^2=C$
$(x-x_1{)}^2+(y-y_1{)}^2+(x-x_2{)}^2-(y-y_2{)}^2+(x-x_3{)}^2+(y-y_3{)}^2=c$
$3x^2-(2x_1+2x_2+2x_3)x+x_1^2+x_2^2+x_3^2+3y^2-2(y_1+y_2+y_3)+y_1^2+y_2^2+y_3^2=c$
$x^2+y^2-\frac{2}{3}(x_1+x_2+x_3)x-\frac{2}{3}(y_1+y_2+y_3)c+(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2-c)=0$
So, center of this circle is $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3}),$
centroid of triangle is a center of that circle.