Question

A point moves so that the sum of the squares of its distances from the four sides of a square is constant ; prove that it always lies on a circle.

Answer 1

Let the sides $AB$ and $BC$ of a square $ABCD$ with $A$ as the origin be along $y$ and $x$ axis respectively and length of side of square be $a$

Let the moving point be $P(h,k)$

Equation of $AB$ is

$x=\mathrm{0.....}\left(i\right)$

Perpendicular distance of $P$ from $\left(i\right)$ $=p_1=h$

Equation of $BC$ is

$y=\mathrm{0......}\left(ii\right)$

Perpendicular distance of $\left(ii\right)$ from $P$ $=p_2=k$

Equation of $CD$ is

$x=a......\left(iii\right)$

Perpendicular dostance of $\left(iii\right)$ from $P$ $=p_3=\frac{h-a}{\sqrt{1^2+0^2}}=h-a$

Equation of $AD$ is

$y=a......\left(iv\right)$

Perpendicular distance of $\left(iv\right)$ from $P$ $p_4=\frac{k-a}{\sqrt{1^2+0^2}}=k-a$

Given : $p_1^2+p_2^2+p_3^2+p_4^2=c$

$\Rightarrow h^2+k^2+{(h-a)}^2+{(k-a)}^2=c\Rightarrow h^2+k^2+h^2+a^2-2ah+k^2+a^2-2ak=c\Rightarrow 2h^2+2k^2-2ah-2ak+2a^2-c=0\Rightarrow h^2+k^2-ah-ak+a^2-\frac{c}{2}=0$

Generalising the equation

$\Rightarrow x^2+y^2-ax-ay+a^2-\frac{c}{2}=0$

Clearly it represents a circle.

Hence proved.