Question

The point $P$ moves in the plane of a regular hexagon such that the sum of the squares of its distances from the vertices of the hexagon is $6a^2$. If the radius of the circumcircle of the hexagon is $r(<a)$, then the locus of $P$ is

• A. a circle of radius $a$
• B. a circle of radius $\sqrt{a^2+r^2}$
• C. a circle of radius $\sqrt{a^2-r^2}$
• D. a circle of radius $ar$

Answer 1

The correct option is C a circle of radius $\sqrt{a^2-r^2}$

Let the center of the circum circle of regular hexagon be origin $O$

From the above figure the vertices are
$A(r\cos 0^{\circ },r\sin 0^{\circ }),B(r\cos {60}^{\circ },r\sin {60}^{\circ }),C(r\cos {120}^{\circ },r\sin {120}^{\circ }),D(r\cos {180}^{\circ },r\sin {180}^{\circ }),E(r\cos {240}^{\circ },r\sin {240}^{\circ }),F(r\cos {300}^{\circ },r\sin {300}^{\circ })$
$\Rightarrow A(r,0),B(\frac{r}{2},\frac{\sqrt{3}r}{2}),C(-\frac{r}{2},\frac{\sqrt{3}r}{2}),D(-r,0),E(-\frac{r}{2},-\frac{\sqrt{3}r}{2}),F(\frac{r}{2},-\frac{\sqrt{3}r}{2})$

If $P=(x,y)$ then,
$\sum (PA{)}^2=6a^2$
$\Rightarrow (x-r{)}^2+y^2+\{{(x-\frac{r}{2})}^2+{(y-\frac{r\sqrt{3}}{2})}^2\}$
$+....+\{{(x-\frac{r}{2})}^2+{(y+\frac{r\sqrt{3}}{2})}^2\}=6a^2$
$\Rightarrow 2(x^2+y^2+r^2)+4(x^2+y^2+\frac{r^2}{4}+\frac{3r^2}{4})=6a^2$
$\Rightarrow x^2+y^2+r^2=a^2$
$\Rightarrow x^2+y^2=(\sqrt{a^2-r^2}{)}^2$