Question

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

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

Answer 1

Answer: (A), (B)

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

There will be two squares possible. The smaller square will have all 4 vertices lying on the smaller circle which touches the four circles. The bigger square will have the vertices lying on the bigger circle which touches the 4 circles.
Refer to the figure.
For the bigger circle,
$OA=\sqrt{2}+1=\frac{d}{2}$
$d=2\sqrt{2}+2$
($d$ is the length of the diagonal of the bigger square)
Side $=\frac{2\sqrt{2}+2}{\sqrt{2}}=2+\sqrt{2}$
For the smaller circle,
$O{A}^{\prime }=\sqrt{2}-1=\frac{d^{\prime }}{2}$
$d^{\prime }=2\sqrt{2}-2$
($d^{\prime }$ is the length of the diagonal of the smaller square)
Side $=\frac{2\sqrt{2}-2}{\sqrt{2}}=2-\sqrt{2}$
Hence, options 'A' and 'B' are correct.