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?

Let us consider the equation of bisecting circle to be "S1"...and the equation of bisected circle be "S2"
S1: x²+y² +4x+22y+l 
S2: x²+y²-2x+8y-m
In order to calculate the value of "l+m"......we first need to calculate the value of common tangent...... 
Let us denote the eqn of common tangent with "L"
The eqn of common tangent will be = S1-S2
So eqn of common tangent is= x²+y²+4x+22y+l - (x²+y²-2x+8y-m) 
Eqn of common tangent = 6x+14y+(l+m)=0
Also the centre of the bisected circle should lie on the common tangent..... 
Centre of bisected circle = (1,-4).                    (-g,-f) 
Putting the coordinates of centre of the circle in eqn of common tangent..... 
6(1)+14(-4)+(l+m)=0
6-56+(l+m)=0
-50 +(l+m)=0
l+m=50
Hence the correct option is (B) 
