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)