The equation of the circles are
$S_1 \equiv x^2+y^2+2 x+3 y+1=0$
and $S_2 \equiv x^2+y^2+4 x+3 y+2=0$
Since, the circles cuts each other at $A$ and $B$ then equation of $A B$ is
$S_1-S_2=0$
$\Rightarrow \quad\left(x^2+y^2+2 x+3 y+1\right)$
$\Rightarrow \quad-\left(x^2+y^2+4 x+3 y+2\right)=0$
$\Rightarrow \quad-2 x-1=0$
$\Rightarrow \quad x=\frac{-1}{2}$
Putting the value $x=-\frac{1}{2}$ in $S_2$, we get
$\frac{1}{4}+y^2-2+3 y+2=0$
$\Rightarrow \quad 4 y^2+12 y+1=0$
$\Rightarrow \quad y=\frac{-12 \pm \sqrt{144-16}}{8}$
$\Rightarrow \quad y=\frac{-12 \pm \sqrt{128}}{8}=\frac{-12 \pm 8 \sqrt{2}}{8}$
$\Rightarrow \quad y=\frac{-3}{2} \pm \sqrt{2}$
So intersection points are
$A\left(-\frac{1}{2},-\frac{3}{2}+\sqrt{2}\right) \text { and }\left(-\frac{1}{2},-\frac{3}{2}-\sqrt{2}\right) \text {. }$
Then equation of circle with diameter $A B$ is
$\left(x+\frac{1}{2}\right)\left(x+\frac{1}{2}\right)+\left(y+\frac{3}{2}+\sqrt{2}\right)\left(y+\frac{3}{2}-\sqrt{2}\right)$ $=0$
$\Rightarrow \quad\left(x+\frac{1}{2}\right)^2+\left(y+\frac{3}{2}\right)^2-2=0$
$\Rightarrow x^2+\frac{1}{4}+x+y^2+\frac{9}{4}+3 y-2=0$
$\Rightarrow \quad x^2+y^2+x+3 y+\frac{1}{2}=0$
$\Rightarrow 2 x^2+2 y^2+2 x+6 y+1=0$