The locus of a point which moves such that the sum of the square of its distance from three vertices of a triangle is constant is a/an circle (b) straight line (c) ellipse (d) none of these

Options

A. circle
B. straight line
C. ellipse
D. none of these

Answer

A

Explanation

Let `A(x_(1), y_(1)), B(x_(2), y_(2))` and `C(x_(3), y_(3))` be the vertices of the triangle ABC, and let P (h, k) be any point on the locus. Then,
`PA^(2)+PB^(2)+PC^(2)=c " " `(constant)
`rArr underset(i=1)overset(3)Sigma(h-x_(i))^(2)+(k-y_(i))^(2)=c`
`rArrh^(2)+k^(2)-(2h)/(3)(underset(i=1)overset(3)Sigma x_( i) )-(2k)/(3)(underset(i=1)overset(3)Sigma y_(i))+(1)/(3)underset(i=1)overset(3)Sigma (x_( i)^(2)+y _(i)^(2))-(c)/(3)=0`
So, locus of (h, k) is
`x^(2)+y^(2)-(2x)/(3)(x_(1)+x_(2)+x_(3))-(2y)/(3)(y_(1)+y_(2)+y_(3))+lambda=0`, where
`lambda=underset(i=1)overset(3)Sigma (x_(i)^(2)+y_(i)^(2))-c=0=` constant.
Clearly, this locus is a circle with centre at
`((x_(1)+x_(2)+x_(3))/(3), (y_(1)+y_(2)+y_(3))/(3))`