Question

lf the circle $x^2+y^2+4x+22y+c=0$ bisects the circumference of the circle $x^2+y^2-2x+8y-d=0$ then $c+d=$

• A. $60$
• B. $50$
• C. $40$
• D. $70$

Answer 1

The correct option is B $50$

If one circle bisects the circumference of another, then the common chord of both circle passes through the center of other circles.
$\therefore$ common chord of given circles is

$6x+14y+c+d=0.$

then this chord passes throught centre of $x^2+y^2-2x+8y-d=0$

centre $(1,-4).$

$\therefore 6\left(1\right)+14(-4)+c+d=0$

$c+d=56-6$

$\Rightarrow c+d=50$