Given that, S1≡x2+y2+6x−2y+k=0 and ‌‌S2≡x2+y2+2x−6y−15=0 Since, S1 bisects S2, then Chord of S2= Diameter of S1 Equation of the chord is S1−S2=0 (x2+y2+6x−2y+k)‌ −(x2+y2+2x−6y−15)‌=0 ⇒‌‌4x+4y+k+15‌=0 Centre of the circle of S2=(−1,3) Since, equation of the chord passes through (−1,3), then ‌4(−1)+4(3)+k+15‌=0 ⇒‌−4+12+k+15‌=0 ⇒‌k‌=−23