Question

A point moves so that the sum of squares of its distances from the points (1,2) and (-2,1) is always 6. Then its locus is

• A. the straight line $y - \frac{3}{2} = -3\left(x+\frac{1}{2}\right)$
• B. a circle with centre $\left(-\frac{1}{2}, \frac{3}{2}\right)$ and radius $\frac{1}{\sqrt{2}}$
• C. a parabola with focus (1,2) and directix pssing through (-2,1)
• D. an ellipse with foci (1,2) and (-2,1)

Answer 1

The Correct Option is B

Let $P$ be any point, whose coordinate is $(h, k)$. Given, $P$ moves, so that the sum of squares of its distances from the points $A(1,2)$ and $B(-2,1)$ is 6 . i.e., $(P A)^{2}+(P B)^{2}=6$ $\Rightarrow (h-1)^{2}+(k-2)^{2}+(h+2)^{2}+(k-1)^{2}=6$ $\Rightarrow h^{2}+1-2\, h+k^{2}+4-4 \,k+h^{2}+4+4 \,h$ $+k^{2}+1-2\, k=6$ $\Rightarrow 2 \,h^{2}+2\, k^{2}+2 \,h-6 \,k+4=0$ $\Rightarrow h^{2}+k^{2}+h-3\, k+2=0$ $\therefore$ Required locus is $x^{2}+y^{2}+x-3\, y+2=0$ Which represent a circle. Whose centre is $\left(\frac{-1}{2}, \frac{3}{2}\right)$ and radius $=\sqrt{\frac{1}{4}+\frac{9}{4}-2}=\sqrt{\frac{5}{2}-2}=\frac{1}{\sqrt{2}}$