Consider a triangle Δ whose two sides lie on the x-axis and the line 𝑥+𝑦+1=0. If the orthocenter of Δ is (1,1), then the equation of the circle passing through the vertices of the triangle Δ is

H (1,1 )=OC
Property : Image of Orthocentre  w.r.t any side lie on circum-circle of the triangle 
Given sides X-axis i.e. $\mathrm{y}=0 \mathrm{\&} \mathrm{x}+\mathrm{y}+1=0$

Image of $\mathrm{H}(1,1)\text{ w.r.t }\mathrm{y}=0\text{ is }\mathrm{P}(1,-1)$
Image of $\mathrm{H}(1,1)\text{ w.r.t }\mathrm{x}+\mathrm{y}+1=0\text{ be }\mathrm{Q}(\alpha ,\beta )$

$\frac{\mathrm{\alpha }-1}{1}=\frac{\mathrm{\beta }-1}{1}=\frac{-2(1+1+1)}{1^2+1^2}=-3,\mathrm{\alpha }=-2,\mathrm{\beta }=-2\Rightarrow \mathrm{Q}(-2,-2)$

Point of Intersection of sides $\mathrm{y}=0 \mathrm{\&} \mathrm{x}+\mathrm{y}+1=0\text{ is }\mathrm{A}(-1,0)\text{. }$

Now, $\mathrm{P}(1,-1),\mathrm{Q}(-2,-2)\mathrm{\&}\mathrm{A}(-1,0)$ lie on circumcircle

slope of $\overrightarrow{\mathrm{PA}}=\frac{1}{-2}$ slope of  $\overleftrightarrow{\mathrm{QA}}=\frac{2}{1}$

Product of slopes = -1.

$\therefore \overrightarrow{\mathrm{PA}}\perp \overrightarrow{\mathrm{QA}}$

$\therefore \overrightarrow{\mathrm{PQ}}$ is diameter of circumcircle . 

circumcircle is $(x-1)(x+2)+(y+1)(y+2)=0$

${\mathrm{x}}^2+\mathrm{x}-2+{\mathrm{y}}^2+3\mathrm{y}+2=0,$

${\mathrm{x}}^2+{\mathrm{y}}^2+\mathrm{x}+3\mathrm{y}=0$