JEE Main 2026 — Apr 5, Shift 1 — Mathematics solution

JEE Main2026Apr 5, Shift 1Mathematics

Let $P$ be a moving point on the circle $x^{2} + y^{2} - 6 x - 8 y + 21 = 0$ . Then, the maximum distance of $P$ from the vertex of the parabola $x^{2} + 6 x + y + 13 = 0$ is equal to :

A
8
B
10
C
12
D
9

Solution

Rewrite the circle as $(x - 3)^{2} + (y - 4)^{2} = 4$ , so its centre is $C (3, 4)$ and radius is $r = 2$ .

Complete the square on the parabola: $(x + 3)^{2} = - (y + 4)$ , so its vertex is $V (- 3, - 4)$ .

Distance from $V$ to $C$ : $(3 - (- 3))^{2} + (4 - (- 4))^{2} = 36 + 64 = 10$ .

The farthest point on the circle from $V$ lies along the line through $V$ and $C$ , at distance $V C + r = 10 + 2 = 12$ .

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