Consider a triangle Δ whose two sides lie on the x -axis and the...

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 $( - 1, 0 ) , ( 1, - 2 ) \text{\text{ and } }$ ( 0, 0 ) Let the equation of the circle is ${x}^{2} + {y}^{2} + 2gx + 2fy = 0 ( 0, 0 ) \to c = 0 ( - 1, 0 ) \to 1 - 2g = 0 ⇒ g = \dfrac{1}{2} ( 1, - 2 ) \to 5 + 1 - 4f = 0 ⇒ f = \dfrac{3}{2}$ 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 $( 1, - 1 ) , \left(- 2, - 2\right) \text{\text{ and } }$ \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)\text{and } \left(iii\right), we get $g = \dfrac{1}{2},$f = $\dfrac{3}{2}, c = 0$ Hence, the equation of the circle is ${x}^{2} + {y}^{2} + x + 3y = 0$