Question

A point moves such that the sum of the squares of its distances from the sides of a square of side unity is equal to $9$. The locus of such a point is a circle

• A. inscribed in the square
• B. circumscribing the square
• C. inside the square
• D. containing the square

Answer 1

The correct option is D containing the square

Let the sides of the square be $x=0,y=0,x=1,y=1$ and the centre of the square being $(0.5,0.5).$
Now for a point $(x,y)$, as per the given condition, $x^2+y^2+(x-1{)}^2+(y-1{)}^2=9$
$\Rightarrow 2x^2+2y^2-2x-2y-7=0$
As we see the equation can be written as $x^2+y^2-x-y-3.5=0$
The centre of the circle is $(0.5,0.5)$ and the radius is $\sqrt{3.5+0.5}=2$
Thus, the square is contained inside the circle.