Circle

$(1,-2)=(\alpha,-\alpha-1)$ $\Rightarrow \alpha=1$ It is clear from question that one of the vertex of triangle is intersection of $x$-axis and $x+y+1=0 \Rightarrow A(-1,0)$ Let vertex B be $(\alpha,-\alpha-1)$ Line $AC \perp BH$ so, $m_{AC} \cdot m_{BH}=-1$ $\Rightarrow O=-\frac{1-\alpha}{\alpha+2} \Rightarrow \alpha=1 \Rightarrow B(1,-2)$ Let vertex $C$ be $(\beta, 0)$ Line $AH \perp BC$ $\therefore m_{AH}-m_{BC}=-1$ $\Rightarrow \frac{1}{2} \cdot \frac{2}{\beta-1}=-1 \Rightarrow \beta=0$ Centroid of $\triangle ABC$ is $\left(0,-\frac{2}{3}\right)$ We know that $G$ (centroid) divides line joining circumcentre $(O)$ and orthocentre $(H)$ in the ratio $1:2$.$\Rightarrow$ Circumcentre is $\left(-\frac{1}{2},-\frac{3}{2}\right)$. Equation of circum circle is (passing through $C(0,0)$ ) is $x^{2}+y^{2}+x+3y=0$