Question

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

• A. x2 + y2 – 3x + y = 0
• B. x2 + y2 + x + 3y = 0
• C. x2 + y2 + 2y – 1 = 0
• D. x2 + y2 + x + y = 0

Answer 1

Answer: (B)

x2 + y2 + x + 3y = 0

Answer 2

Given that the mirror image of the orthocenter lies on the circumcircle:

The image of the point (1, 1) reflected over the x-axis is (1, -1). The image of the point (1, 1) reflected over the line $x+y+1$=0 is (-2, -2).

Therefore, the circle passing through both (1, -1) and (-2, -2) is determined.

Thus, the circle represented by the equation x2 + y2 + x + 3y = 0 satisfies this condition.
Hence the correct option is B