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

$\text{Two sides lie on: } y = 0 \ \text{ and } \ x + y + 1 = 0$ $\text{Orthocentre } H = (1, 1)$ $\text{To find: circle through the 3 vertices (circumcircle)}$ $\text{Options: } x^2 + y^2 + bx + cy + d = 0$ $\text{Sides: } y = 0 \ \text{ and } \ x + y + 1 = 0$ $A = \{y=0\} \cap \{x+y+1=0\} = (-1, 0)$ $AB \subset y=0 \Rightarrow \text{altitude from } C \text{ is vertical, through } H(1,1)$ $\Rightarrow x_C = 1$ $C = (1, y_C) \text{ on } x+y+1=0 \Rightarrow 1 + y_C + 1 = 0$ $C = (1, -2)$ $AC \subset x+y+1=0,\ \text{slope} = -1 \Rightarrow \text{altitude from } B: \text{slope } 1$ $y = x,\qquad B = (0, 0)$ $\text{Let circle: } x^2 + y^2 + Dx + Ey + F = 0$ $B(0,0):\quad F = 0$ $A(-1,0):\quad 1 - D + F = 0 \Rightarrow D = 1$ $C(1,-2):\quad 1 + 4 + D - 2E + F = 0$ $5 + 1 - 2E = 0 \Rightarrow E = 3$ $D = 1,\qquad E = 3,\qquad F = 0$ $\boxed{x^2 + y^2 + x + 3y = 0} \quad \Rightarrow \text{option (2)}$ $\text{centre } \left(-\dfrac{1}{2}, -\dfrac{3}{2}\right),\quad R^2 = \dfrac{D^2+E^2}{4} - F = \dfrac{5}{2}$ $\text{every vertex is at distance } R \ \Rightarrow \text{correct}$