JEE Main 2026 — Mathematics Coordinate Geometry

The equation of the given circle is $x^2 + y^2 - 6x - 8y + 21 = 0$x2+y2−6x−8y+21=0.

The center of the circle is $C(3, 4)$C(3,4) and its radius is $r = \sqrt{3^2 + 4^2 - 21} = \sqrt{4} = 2$r=32+42−21​=4​=2.

The equation of the given parabola is $x^2 + 6x + y + 13 = 0$x2+6x+y+13=0.

Rewriting the equation by completing the square, we get $(x + 3)^2 = -(y + 4)$(x+3)2=−(y+4).

The vertex of the parabola is $V(-3, -4)$V(−3,−4).

The maximum distance of a moving point $P$P on the circle from the vertex $V$V is given by $CV + r$CV+r.

The distance between the center $C(3, 4)$C(3,4) and the vertex $V(-3, -4)$V(−3,−4) is $CV = \sqrt{(3 - (-3))^2 + (4 - (-4))^2} = \sqrt{6^2 + 8^2} = \sqrt{100} = 10$CV=(3−(−3))2+(4−(−4))2​=62+82​=100​=10.

Therefore, the maximum distance is $10 + 2 = 12$10+2=12.

Answer: $12$12