Question

A point $P$ moves such that the sum of the square of its distances from the sides of a square of side unity is $9$. Then the locus of $P$ is

• A. $x^2+y^2=3$
• B. $x^2-y^2=4$
• C. $x^2+y^2=2$
• D. $x^2+y^2-x-y-\frac{7}{2}=0$

Answer 1

The correct option is D

$x^2+y^2-x-y-\frac{7}{2}=0$

Explanation for the correct option:

Find the distance of the point from the side of the square:

Assume the square with sides as $x=0,y=0,x=1,y=1$.

Find the distance using the given conditions.

$$\begin{gathered} x^2+y^2+{\left(x-1\right)}^2+{\left(y-1\right)}^2=9 \\ \Rightarrow x^2+y^2+x^2-2x+1+y^2-2y+1-9=0 \\ \Rightarrow 2x^2+2y^2-2x-2y-7=0 \\ \Rightarrow x^2+y^2-x-y-3.5=0 \end{gathered}$$

Hence, option (D) is the correct answer.