AP EAMCET Engineering 20 Apr 2019 Shift 2 Paper

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