If x–y +1 = 0 meets the circle $x^2+y^2+y-1=0$at A and B then the equation of the circle with AB as diameter is

$$\begin{gathered} \mathrm{L}: \mathrm{x}-\mathrm{y}+1=0 \\ \mathrm{S}: {\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{y}-1=0 \end{gathered}$$

Equation of circle is 

$$\begin{gathered} \mathrm{S}+\mathrm{\lambda }\left(\mathrm{L}\right)=0 \\ \left({\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{y}-1\right)+\mathrm{\lambda }\left(\mathrm{x}-\mathrm{y}+1\right)=0 \\ {\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{\lambda x}+\mathrm{y}\left(1-\mathrm{\lambda }\right)-1+\mathrm{\lambda }=0 \\ \mathrm{C}=\left(\frac{-\mathrm{\lambda }}{2}, \frac{\mathrm{\lambda }-1}{2}\right) \end{gathered}$$

It is lies on $\mathrm{x}-\mathrm{y}+1=0$

$$\begin{gathered} \frac{-\mathrm{\lambda }}{2}-\left(\frac{\mathrm{\lambda }-1}{2}\right)+1=0 \\ \Rightarrow -\mathrm{\lambda }-\mathrm{\lambda }+1+2=0 \\ \Rightarrow 2\mathrm{\lambda }=3 \\ \Rightarrow \mathrm{\lambda }=\raisebox{1ex}{$3$}\!\left/ \!\raisebox{-1ex}{$2$}\right. \\ So,required circle is \left({\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{y}-1\right)+\frac{3}{2}\left(\mathrm{x}-\mathrm{y}+1\right)=0 \\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2+2\mathrm{y}-2+3\mathrm{x}-3\mathrm{y}+3=0 \\ \Rightarrow 2{\mathrm{x}}^2+2{\mathrm{y}}^2+3\mathrm{x}-\mathrm{y}+1=0 \end{gathered}$$