JEE Main 2026 — Circle Question with 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: $12$