Let ABCD be a square such that vertices A,B,C,D lie on circles x2+y2−2x−2y+1=0, x2+y2+2x−2y+1=0, x2+y2+2x+2y+1=0 and x2+y2−2x+2y+1=0respectively with centre of square being origin and sides are parallel to coordinate axes. The length of the side of such a square can be
2+√2
3−√3
3+√3
2−√2
A
3−√3
B
3+√3
C
2+√2
D
2−√2
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Solution
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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.
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Q1
Let ABCD be a square such that vertices A,B,C,D lie on circles x2+y2−2x−2y+1=0, x2+y2+2x−2y+1=0, x2+y2+2x+2y+1=0 and x2+y2−2x+2y+1=0respectively with centre of square being origin and sides are parallel to coordinate axes. The length of the side of such a square can be
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Q2
The equation of the circle in the first quadrant touching each coordinate axes at a distance of one unit from the origin is
(a) x2 + y2 – 2x – 2y + 1 = 0
(b) x2 + y2 – 2x – 2y – 1 = 0
(c) x2 + y2 – 2x – 2y = 0
(d) x2 + y2 – 2x + 2y – 1 = 0