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Physics Guru is an online education and competitive examinations preparation portal founded by Manish Verma, an IIT Madras alumnus, providing learning resources for competitive examinations.

The emphasis is on conceptual understanding, analytical thinking and problem solving rather than simply memorising formulas and standard methods.

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Consider a triangle $\Delta$ whose two sides ...

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://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)