The circle $x^2+y^2=4$ cuts the circle $x^2+y^2+2x+3y-5=0$ in A and B. Then the equation of the circle on AB as diameter, is

The equation of the circle on AB as diameter is 

$\left(x^2+y^2-4\right)+\lambda (2x+3y-1)=0\left[U\mathrm{sing}{S}_1+\lambda \left(S_2-S_1\right)=0\right]$

$\text{ or, }{x}^2+y^2+2\lambda x+3\lambda y-(\lambda +4)=0$

The equation of $AB$ is

$2x+3y-1=0$

Ii $AB$ is a diameter of (i), then centre lies on $AB$.

$-2\lambda -\frac{9\lambda }{2}-1=0\Rightarrow \lambda =-\frac{2}{13}$

$13x^2+13y^2-4x-6y-50=0$