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

Solution

Reflection of orthocentre about side of a triangle lies on the circumcircle. (Refer https://123iitjee.manishverma.site/for-a-triangle-show-that-the-point-where-altitude/)

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)