AP EAMCET Engineering 2013 Solved Paper

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