Question

Let $ABCD$ be a square such that vertices $A,B,C,D$ lie on circles $x^2+y^2-2x-2y=0$, $x^2+y^2-2x-2y+1=0$, $x^2+y^2+2x+2y+1=0$ and $x^2+y^2-2x+2y+1=0$ respectively with center of square being origin and sides are parallel to coordinate axes. The length of side of such square can be

• A. $2-\sqrt{2}$
• B. $2+\sqrt{2}$
• C. $3-\sqrt{3}$
• D. $3+\sqrt{3}$

Answer 1

Answer: (A), (B)

$2+\sqrt{2}$
$2-\sqrt{2}$

$OA=\sqrt{2}+1=\frac{d}{2}$
$d=2\sqrt{2}+2$
$Side=\frac{2\sqrt{2}+2}{\sqrt{2}}=2+\sqrt{2}$
$O{A}^{\prime }=\sqrt{2}-1=\frac{d^{\prime }}{2}$
$d^{\prime }=2\sqrt{2}-2$
$Side=\frac{2\sqrt{2}-2}{\sqrt{2}}=2-\sqrt{2}$
Hence, options 'A' and 'B' are correct.