Q35 / 75 · JEE Main 2026 (05 April Shift 1) · Mathematics
Let $P$ be a moving point on the circle $x^2 + y^2 - 6x - 8y + 21 = 0$. Then, the maximum distance of $P$ from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:
- A$8$
- B$10$
- Correct answerC$12$
- D$9$
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 $r = \sqrt{3^2 + 4^2 - 21} = \sqrt{4} = 2$.
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 $CV = \sqrt{(3 - (-3))^2 + (4 - (-4))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$.
Therefore, the maximum distance is $10 + 2 = 12$.
Answer: $12$
The center of the circle is $C(3, 4)$ and its radius is $r = \sqrt{3^2 + 4^2 - 21} = \sqrt{4} = 2$.
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 $CV = \sqrt{(3 - (-3))^2 + (4 - (-4))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$.
Therefore, the maximum distance is $10 + 2 = 12$.
Answer: $12$
Hence, the correct answer is option: 'C'.