JEE Main20265 April 2026Evening ShiftMathematicsParabolaActual
Let the point P be the vertex of the parabola y = x^2 - 6x + 12 . If a line passing through the point P intersects the circle x^2 + y^2 - 2x - 4y + 3 = 0 at the points R and S , then the maximum value of (PR + PS)^2 is :
Options
- A10
- B20
- C25
- D5
Correct answer
B. 20
Step-by-step solution
The equation of the parabola is y = x^2 - 6x + 12 , which can be rewritten as y - 3 = (x - 3)^2 . The vertex of the parabola is P(3, 3) . The equation of the circle is x^2 + y^2 - 2x - 4y + 3 = 0 , which can be rewritten as (x - 1)^2 + (y - 2)^2 = 2 . The center of the circle is C(1, 2) and its radius is r = 2 . The distance between the point P and the center C is: PC = (3 - 1)^2 + (3 - 2)^2 = 4 + 1 = 5 Since PC = 5 > r = 2 , the point P lies outside the circle. Let the line through P intersect the circle at R and