UPSC NDA I 2015 Mathematics Solved Paper

Let ‘r’ be the radius of the circle and (x, y) be the coordinates of the circumcenter.
Therefore consider the following equations.
(x+2)2+(y−3)2−r2
(x−2)2+(y−1)2=r2
(x−1)2+(y−2)2=r2
8x=(2y−4)(2)
4x=2y−4
2x=y−2
2x−y=−2.......(1)
(2x+1)(3)=(2y−5)
6x+3=2y−5
6x−2y=−8
3x−y=−4...(2)
Hence the circumcenter lies at the intersection of the following lines
2x−y=−2...(1)
3x−y=−4...(2)
Solving 1 and 2,we get the circumcenter as
C=(−2,−2)