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 locusis

• A. The straight line $Y-\frac{3}{2}=-3(x+\frac{1}{2})$
• B. A circle with centre $(-\frac{1}{2},\frac{3}{2})$
• C. A parabola with focus $(1,2)$ and directly passing through $(-2,1)$
• D. An ellipse with foci $(1,2)$ and $(-21)$

Answer 1

The correct option is B A circle with centre $(-\frac{1}{2},\frac{3}{2})$

$(x-1{)}^2+(y-2{)}^2+(x+2{)}^2+(y-1{)}^2=6$

$⟹2x^2+2y^2+2x-6y+10=6$

$⟹x^2+y^2+x-3y=-2$

$⟹{(x+\frac{1}{2})}^2+{(y-\frac{3}{2})}^2=-2+\frac{1}{4}+\frac{9}{4}$

$⟹{(x+\frac{1}{2})}^2+{(y-\frac{3}{2})}^2={\left(\frac{1}{\sqrt{2}}\right)}^2$

This is a circle with its centre at $(-\frac{1}{2},\frac{3}{2})$

So, the answer is option B