JEE MainMathematicsStraight Lines
In a triangle, the orthocentre is (2, 3) , the circumcentre is (1, 2) , and the equation of one of its sides is x + y = 2 . The square of the circumradius of the triangle is
Options
- A8
- B2
- C4
- D32
Correct answer
A. 8
Step-by-step solution
Let the given side of the triangle be BC with equation x + y - 2 = 0 . A known property of a triangle is that the reflection of its orthocentre across any side lies on its circumcircle. Let the orthocentre be H(2, 3) and its reflection across the line x + y - 2 = 0 be H'(x', y') . Using the reflection formula x' - x₁ a = y' - y₁ b = -2 ax₁ + by₁ + c a^2 + b^2 , we get: x' - 2 1 = y' - 3 1 = -2 2 + 3 - 2 1^2 + 1^2 x' - 2 1 = y' - 3 1 = -2 ( 3 2 ) = -3 This gives x' - 2 = -3 x' = -1 and y' - 3 = -3 y' = 0 . So, the p