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

Question

The locus of a point which moves such that the sum of the square of its distances from the three vertices of a triangle is constant, is a circle whose centre is at the (A) incentre of the triangle (B) centroid of the triangle (C) orthocentre of the triangle (D) none of these

Tardigrade Admin · Accepted Answer

Let a triangle has its three vertices as

(0, 0), (a, 0), (0, b)

.
We have the moving point

(h, k)

such that

h^{2} + k^{2} + (h - a)^{2} + k^{2} + h^{2} + (k - b)^{2} = c

\Rightarrow 3 h^{2} + 3 k^{2} - 2 ah - 2 bk + a^{2} + b^{2} = c

Therefore,

3 x^{2} + 3 y^{2} - 2 a x - 2 b y + a^{2} + b^{2} = c

Its centre is

(\frac{a}{3}, \frac{b}{3})

, which is centroid of triangle.

Q. The locus of a point which moves such that the sum of the square of its distances from the three vertices of a triangle is constant, is a circle whose centre is at the

incentre of the triangle

centroid of the triangle

orthocentre of the triangle

none of these

Let a triangle has its three vertices as (0,0),(a,0),(0,b). We have the moving point (h,k) such that h2+k2+(h−a)2+k2+h2+(k−b)2=c ⇒3h2+3k2−2ah−2bk+a2+b2=c Therefore, 3x2+3y2−2ax−2by+a2+b2=c Its centre is (3a​,3b​), which is centroid of triangle.