JEE Advanced 2014 Paper 1

Given Circles
x2+y2−2x−15=0
x2+y2−1=0
Radical axis x+7=0...(1)
Center of circle lies on (1)
Let the center be (-7,k)
Let equation be x2+y2+14x−2ky+c=0
Orthogonality gives
−14=c−15⇒c=1....(2)
(0,1)→1−2k+1=0⇒k=1
Hence radius=√72+k2−c=√49+1−1=7
Alternate solution
Given circle x2+y2−2x−15=0
x2+y2−1=0
Let equation of circle x2+y2+2gx+2fy+c=0
Circle passes through (0,1)
⇒1+2f+c=0
Applying condition of orthogonality
−2g=c−15,0=c−1
⇒c=1,g=7,f=−1
r=√49+1−1=7;center (−7,1)