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

(A) $x^2+y^2-3x+y=0$

(B) $x^2+y^2+x+3y=0$

(C) $x^2+y^2+2y-1=0$

(D) $x^2+y^2+x+y=0$

(B) $x^2+y^2+x+3y=0$

(C) $x^2+y^2+2y-1=0$

(D) $x^2+y^2+x+y=0$

*Solution*

Reflection of orthocentre about side of a triangle lies on the circumcircle. (Refer https://www.123iitjee.com/2022/03/for-triangle-show-that-point-where.html )

Reflection of orthocentre (1, 1) about side of triangle lying on x-axis is the point (1, -1). Out of 4 options, only option (B) satisfies this.

Let (h, k) be the reflection of orthocentre (1, 1) about side of triangle lying on the line x+y+1=0.

Then, the point $\left( {\frac{{h + 1}}{2},\frac{{k + 1}}{2}} \right)$ lies on the line x+y+1=0.

$\Rightarrow \frac{{h + 1}}{2} + \frac{{k + 1}}{2} + 1 = 0$

Or, h+k+4=0 ....(*)

Also, $\frac{{k - 1}}{{h - 1}} \times - 1 = - 1$

Or h = k

From (*), h = k = -2

As expected, (-2, -2) satisfies option (B).

Answer: (B)