Thank you for reporting, we will resolve it shortly
Q.
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
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 $