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?

Write as $S_1 :x^2+y^2=4$ and $S_2 :x^2+y^2=-2x-3y+5$
Now to find the equation of AB: $S_1=S_2$
$4=-2x-3y+5\Longrightarrow \text{ the equation of line AB: }y=\frac 1 3 -\frac{2x}{3}\\
\text{let's find A,B: } (\frac 1 3 -\frac{2x}{3})^2+x^2=4\Longrightarrow  x_{1,2}=\frac{2}{13}\pm\frac{3\sqrt{51}}{13}\\
(\frac{2}{13}+\frac{3\sqrt{51}}{13})^2+y^2=4\Longrightarrow  y_{1,2}=\pm\frac{1}{13}\sqrt{3(71-4\sqrt{51})}\\
\text{now find the center $a,b$ using } a=\frac{x_1+x_2}{2}\text{ and } b=\frac{y_1+y_2}{2}\\
 (x-a)^2+(y-b)^2=R^2\\
$
