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. 30
• B. 50
• C. 40
• D. 56
• E. 52

Answer 1

Answer: (B)

Given that, $S_1\equiv x^2+y^2+4x+22y+c=0$ bisects the circumference of the circle $S_2\equiv x^2+y^2-2x+8y-d=0$
The common chord of the given circle is $S_1-S_2=0$
$\Rightarrow$ $x^2+y^2+4x+22y+c-x^2-y^2$ $+2x-8y+d=0$
$\Rightarrow$ $6x+14y+c+d=0$ ...(i)
So, Eq. (i) passes through the centre of the second circle, ie, $(1,-4)$ .
$\therefore$ $6-56+c-4-d=0$
$\Rightarrow$ $c+d=50$