JEE Main 2026 · 5 Apr · Shift 1 · Mathematics Q10

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MathematicsSingle correct+4 / -1Question 10 of 75

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:

Answer and worked solution

Official answer: C

Circle. Centre $(3,4)$ , radius $\sqrt{9+16-21}=2$ .

Parabola. $(x+3)^2=-(y+4)$ , so the vertex is $(-3,-4)$ .

Maximum. The centre is $\sqrt{36+64}=10$ from the vertex, so the farthest point of the circle is $10+2=12$ away.

(C)

Topics:Equation of a circle Parabola

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JEE Main 2026 · 5 Apr · Shift 1, Mathematics question 10