Question

If 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 $c+d$ is equal to :

• A. 60
• B. 50
• C. 40
• D. 30

Answer 1

Answer: (B)

Given that, circle $x^2+y^2+4x+22y+c=0$
bisects the circumference of the circle
$x^2+y^2-2x+8y-d=0$
$\therefore$ The common chord of the given circle is
$S_1-S_2=0$
$\Rightarrow x^2+y^2+4x+22v+c-x^2-y^2$
$+2x-8y+d=0$
$\Rightarrow 6x+14y+c+d=0$
So, Eq. (i) passes through the centre of the second circle i.e., $(1,-4)$
$\therefore 6-56+c+d=0$
$\Rightarrow c+d=50$
Note : If $S_1$ and $S_2$ are the equations of two circles,
then equation of common chord is $S_1-S_2=0$.