JEE Main Mathematics Question (2026) — Solution

Step-by-step solution

The equation of the given circle is x^2 + y^2 - 6x - 8y + 21 = 0. The center of the circle is C(3, 4) and its radius is r = 3^2 + 4^2 - 21 = 4 = 2. The equation of the given parabola is x^2 + 6x + y + 13 = 0. Rewriting the equation by completing the square, we get (x + 3)^2 = -(y + 4). The vertex of the parabola is V(-3, -4). The maximum distance of a moving point P on the circle from the vertex V is given by CV + r. The distance between the center C(3, 4) and the vertex V(-3, -4) is CV = (3 - (-3))^2 + (4 - (-4))^2 = 6^2 + 8^2 = 100 = 10. Therefore, the maximum distance is 10 + 2 = 12. Answer: 12