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

Let $A$ is the point of intersection of the given lines $y=0, x+y+1=0$

Their point of intersection $A$ is $\left(-1,0\right)$

Let $B$ is the other point on the line $x+y+1=0$,

Now, the perpendicular from $B$ on the $x$-axis will pass through the orthocenter.

So, the equation of perpendicular from $B$ on the $x$-axis is $x=1$

Thus, the coordinates of $B$ are $\left(1,-2\right)$

Let the third vertex is $\left(h,k\right)$

Slope of the line perpendicular to $x+y+1=0$ is $1$

So, the perpendicular from $\left(h,k\right)$ to line $x+y+1=0$ has slope $1$ and passes through the orthocenter $\left(1,1\right)$

Thus, its equation is $y=x$

Now, $y=x$ and $x$-axis intersects at the origin.

Hence, the vertices of the triangle are $(-1,0), (1,-2) \& (0,0)$

Let the equation of the circle is $x^2+y^2+2gx+2fy=0$

$(0,0)\to c=0$

$(-1,0)\to 1-2g=0\Rightarrow g=\frac{1}{2}$

$(1,-2)\to 5+1-4f=0\Rightarrow f=\frac{3}{2}$

Hence, the equation of the circumcircle is $x^2+y^2+x+3y=0$.

Aliter

We know, the image of the orthocentre about the sides of a triangle lies on circumcircle.

The image of $(1,1)$ about $y=0$ is $(1,-1)$

The image of $\left(1,1\right)$ about $x+y+1=0$ is $(-2,-2)$

& intersection of $y=0, x+y+1=0$ is $(-1,0).$

Let the equation of circle is 

$x^2+y^2+2gx+2fy+c=0$

All the three points $(1,-1), \left(-2,-2\right) \& \left(-1,0\right)$ lies on this circle

$(1,-1)\to 2+2g-2f+c=0 ...\left(i\right)$

$(-1,0)\to 1-2g+c=0 ...\left(ii\right)$

$(-2,-2)\to 8-4g-4f+c=0 ...\left(iii\right)$

By using the equations $\left(i\right), \left(ii\right) \& \left(iii\right)$, we get

$g=\frac{1}{2}, f=\frac{3}{2}, c=0$

Hence, the equation of the circle is $x^2+y^2+x+3y=0$