Consider the circles. S1=x2+y2+6x−2y+k=0 And S2=x2+y2+2x−6y−15=0 The circle S1 bisects the circumference of the circle S2, then the common circles S1 and S2 will pass through the circle S2(−1,3) The equation of common chord of the circle S1 and S2 is 4x+4y+(k+15)=0 since the chord passes through point (−1,3). Thus, −4+12+k+15=0 k=−23