If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ the...

Question

Problem : If the circle $x^2+y^2+4x+22y+c=0$ bisects the circmuference of the circle $x^2+y^2-2x+8y-d=0$ then c +d equals

(a) 60

(b) 50

(c) 40

(d) 30

Solution : Equation of common chord of the circles is given by $S-S'=0$ where $S = x^2+y^2+4x+22y+c=0 ; S' = x^2+y^2-2x+8y-d=0$

$\Rightarrow 4x+22y+c-(2x+8y-d)=0$

$\Rightarrow 4x+22y+c-2x-8y+d =0$ $\Rightarrow c+d = -(2x+14y)$

Now how to get the value of c +d , please suggest thanks..

Answer 1

The Common chord has equation $6x+14y+c+d=0$ and it must pass through the center $(1,-4)$ of $S'$. Thus $c+d=50$.

Answer 2

you had an error: $$S-S'=4x+22y+c-(-2x+8y-d)\\=6x+14y+c+d=0$$ so $$c+d=-6x-14y$$

because $S$ besects $S'$, so the center of $S'$ is on the line $S-S'$. The center is $(1,-4)$, so $$c+d=-6*1-14*(-4)=50$$