Let ABCD be a square such that vertices A,B,C,D lie on circles $x^{2} + y^{2} - 2 x - 2 y + 1 = 0$ , $x^{2} + y^{2} + 2 x - 2 y + 1 = 0$ , $x^{2} + y^{2} + 2 x + 2 y + 1 = 0$ and $x^{2} + y^{2} - 2 x + 2 y + 1 = 0$ 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

Question

Let ABCD be a square such that vertices A,B,C,D lie on circles

x^{2} + y^{2} - 2 x - 2 y + 1 = 0

,

x^{2} + y^{2} + 2 x - 2 y + 1 = 0

,

x^{2} + y^{2} + 2 x + 2 y + 1 = 0

and

x^{2} + y^{2} - 2 x + 2 y + 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

Toppr Admin · Accepted Answer

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-xA0-

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.