Question

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

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

Answer 1

Answer: (B)

a circle with centre $(-\frac{1}{2},\frac{3}{2})$ and radius $\frac{1}{\sqrt{2}}$

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