The common chord of the given circle is S1−S2=0 ⇒ (x2+y2+4x−6y+c)−(x2+y2−6x+4y−12)=0 ⇒ 10x−10y+c+12=0 ......(i) Since, circle x2+y2+4x+6y+c=0 bisects the circumference of the circle. x2+y2−6x+4y−12=0 Therefore, Eq. (i) passes through the centre of second circle i.e., (3,−2) ∴10(3)−10(−2)+c+12=0 ⇒30+20+c+12=0 ⇒c=−62