Question

If a point $P$ moves such that the sum of the squares of its distances from the three vertices of a $△ABC$ is constant, then the locus of the point $P$ is a circle whose center is

• A. circumcentre of $\Delta ABC$.
• B. orthocentre of $\Delta ABC$.
• C. incentre of $\Delta ABC$.
• D. centroid of $\Delta ABC$.

Answer 1

Answer: (D)

centroid of $\Delta ABC$.

Let point $P$ be $(h,k)$. Let $(x_1,y_1),(x_2,y_2)$ and $(x_3,y_3)$ be the coordinates of the vertices
The sum of the squares of the distances from the three vertices is constant.
$\Rightarrow (h-x_1{)}^2+(k-y_1{)}^2+(h-x_2{)}^2+(k-y_2{)}^2+(h-x_3{)}^2+(k-y_3{)}^2=C$
$\Rightarrow h^2+x_1^2-2h{x}_1+k^2+y_1^2-2y_{1}k+h^2+x_2^2-2h{x}_2+k^2+y_2^2-2k{y}_2+h^2+x_3^2-2h{x}_3+k^2+y_3^2-2k{y}_3=C$
$\Rightarrow h^2+k^2-\frac{2h(x_1+x_2+x_3)}{3}-\frac{2k(y_1+y_2+y_3)}{3}=C-(y_1{}^2+y_2{}^2+y_3{}^2+x_1{}^2+x_2{}^2+x_3{}^2)$
Hence, the locus is a circle with coordinates of the center as $(\frac{x_1+x_2+x_3}{3},\frac{y_1+y_2+y_3}{3})$

These are the coordinates of the centroid of $△ABC$.