Question

A point moves in such a manner that the sum of the squares of its distances from the vertices of a triangle is constant. Then, the locus of the point is

• A. a hyperbola
• B. a parabola
• C. an ellipse
• D. a circle

Answer 1

The correct option is D a circle

Let $A(x_1,y_1),B(x_2,y_2)\text{and}C(x_3,y_3)$ be the vertices of a $\Delta ABC$.
Let $P(h,k)$ be a variable point such that
$P{A}^2+P{B}^2+P{C}^2={\lambda }^2\text{(constant)}$
$\Rightarrow (x_1-h{)}^2+(y_1-k{)}^2+(x_2-h{)}^2$+$(y_2-k{)}^2+(x_3-h{)}^2+(y_3-k{)}^2={\lambda }^2$
$\Rightarrow 3h^2+3k^2-2h(x_1+x_2+x_3)-2k(y_1+y_2+y_3)$
$+x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2-{\lambda }^2=0$
Hence, the locus of $(h,k)$ is
$3x^2+3y^2-2(x_1+x_2+x_3)x-2(y_1+y_2+y_3)y$
$+x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2-{\lambda }^2=0$
$\Rightarrow x^2+y^2-2\left(\frac{x_1+x_2+x_3}{3}\right)x-2\left(\frac{y_1+y_3+y_3}{3}\right)y$
$+\frac{1}{3}(x_1^2+x_2^2+x_3^2+y_1^2+y_2^2+y_3^2-{\lambda }^2)=0$
Clearly, it represents a circle.