<s>[INST] 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... 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..
 [/INST] 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
$$c+d=-6*1-14*(-4)=50$$
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