If the circle $x^2 + y^2 + 4x + 22y + c = 0$bisects the circumference of the circle  $x2 + y^2-2x + 8y - d = 0$ (where c and d are greater than zero). Then maximum value of cd is

One circle bisects the circumference of 2^nd circle, is common chord of both circles $6x+14y+\left(c+d\right)=0$ becomes a diameter of 2^nd circle.

Sub (1,-4) in $6x+14y+\left(c+d\right)=0$

6(1)-4(14)+c+d=0

Therefore c + d = 50
since A.M $\ge$ G.M
$\begin{array}{l}\frac{\mathrm{c}+\mathrm{d}}{2}\ge \sqrt{\mathrm{cd}}\\ \mathrm{cd}\le 625\end{array}$