Thank you for reporting, we will resolve it shortly
Q.
The circle $x^2+y^2=4$ cuts the circle $x^2+y^2+2 x+3 y-5=0$ in $A$ & $B$. Then the equation of the circle on $AB$ as a diameter is:
Conic Sections
Solution:
Common chord of given circle $2 x+3 y-1=0$ family of circle passing through point of intersection of given circle
$ \left(x^2+y^2+2 x+3 y-5\right)+\lambda\left(x^2+y^2-4\right)=0$
$ (\lambda+1) x^2+(\lambda+1) y^2+2 x+3 y-(4 \lambda+5)=0 $
$ x^2+y^2+\frac{2 x}{\lambda+1}+\frac{3}{\lambda+1} y-\frac{(4 \lambda+5)}{\lambda+1}=0$
centre $\left(-\frac{1}{\lambda+1}, \frac{-3}{2(\lambda+1)}\right)$
This centre lies on
$ 2\left(-\frac{1}{\lambda+1}\right)+3\left(\frac{-3}{2(\lambda+1)}\right)-1=0 $
$ -4-9-2 \lambda-2=0$
$ \Rightarrow 2 \lambda=-15 \Rightarrow \lambda=-15 / 2 $
$ \left(-\frac{15}{2}+1\right) x^2+\left(-\frac{15}{2}+1\right) y^2+2 x+3 y-\left(-4 \times \frac{15}{2}+5\right)=0 $
$ \Rightarrow-\frac{13 x^2}{2}-\frac{13 y^2}{2}+2 x+3 y+25=0 $
$ \Rightarrow 13\left(x^2+y^2\right)-4 x-6 y-50=0$
