Family of circles with $AB$ as diameter The circle $S_1 :x^2+y^2-4=0$ cuts the circle $S_2 :x^2+y^2+2x+3y-5=0$ in A and B. Then find the equation of circle with $AB$ as diameter. 
Answer is $13(x^2+y^2)-4x-6y-50=0$
Equation of AB will $S_2-S_1$ and then I used the fact that centre of one of the circles in the family of circles $S_1+kS_2=0$ must lie on $AB$ but I am not getting the correct answer. Any suggestions?

Your reasoning is correct, so I suspect you just made an algebra error.
As you say, the line $AB$ has equation $S_1 - S_2=0$, which is
$$
2x +3y - 1 = 0
$$
Any circle $C_k$ through $A$ and $B$ has equation $S_1 + kS_2=0$ for some $k$. Doing the algebra tells us that $C_k$ has equation
$$
(1+k)x^2 + (1+k)y^2 +2x +3y -5 - 4k = 0
$$
The center of $C_k$ is at the point
$$
P_k = - \left( \frac{2}{2(1+k)},  \frac{3}{2(1+k)} \right)
$$
If $C_k$ has diameter $AB$, then $P_k$ lies on $AB$, so
$$
-\frac{4}{2(1+k)} - \frac{9}{2(1+k)} = 1
$$
This gives $1+k = -13/2$ and $k = -15/2$. Substituting back in the equation of $C_k$ gives
$$
13(x^2+y^2)-4x-6y-50=0
$$
