Circle bisecting the circumference of another circle If the circle $x^2+y^2+4x+22y+l=0$ bisects the circumference of the circle $x^2+y^2-2x+8y-m=0$,then $l+m$ is equal to
(A)$\ 60$
(B)$\ 50$
(C)$\ 46$
(D)$\ 40$
I don't know the condition when one circle intersects the circumference of other circle. So could not solve this question. Can someone help me in this question?

Circle $C_1: x^2+y^2+4x+22y+l=0$ has its center $(-2, -11)$ & a radius $\sqrt{(-2)^2+(-11)^2-l}=\sqrt{125-l}$  
Similarly, circle $C_2: x^2+y^2-2x+8y-m=0$ has its center $(1, -4)$ & a radius $\sqrt{(1)^2+(-4)^2-(-m)}=\sqrt{m+17}$ 
Now, solving the equations of circles $C_1$ & $C_2$ by substituting the value of $(x^2+y^2)$ from $C_2$ into $C_1$, we get the $\color{blue}{\text{equation of common chord}}$ as follows  $$(2x-8y+m)+4x+22y+l=0$$ $$\color{blue}{6x+14y+(l+m)=0}\tag 1$$ 
Now, since the circumference of circle $C_2$ is bisected by the circle $C_1$ hence the center $(1, -4)$ of circle $C_2$ must lie on the common chord or in other words, the common chord must pass through the center of circle $C_2$
Now, satisfying the above equation of common chord by center point $(1, -4)$ as follows
$$6(1)+14(-4)+(l+m)=0$$ $$6-56+l+m=0$$ $$\bbox[5px, border:2px solid #C0A000]{l+m=50}$$
