Mathematics · Circles

JEE Main 2026 — 5 April, Morning Shift — Question 33

Let P be a moving point on the circle $x²+y²-6x-8y+21=0.$ Then the maximum distance of P from the vertex of the parabola $x²+6x+y+13=0$ is equal to:

Option A:
$8$
Option B:
$10$
Option C:
$12$
Correct
Option D:
$9$

Answer: C

Step-by-step solution

Centre : $\mathrm{C}(3,4) \& \mathrm{r}=2$ Parabola $(\mathrm{x}+3)^{2}=-(\mathrm{y}+4)$ Vertex is $\mathrm{A}(-3,-4)$ $\mathrm{AP}_{\text {max }}=\mathrm{AC}+\mathrm{r}=\sqrt{36+64}+2=12$

Answer key and solution verified before publishing.

Practise Circles

Start with this question, then two more from the same chapter — with a tutor that explains every step. Free.

Solve a similar one free

Exam: JEE Main 2026
Paper: 5 April, Morning Shift
Subject: Mathematics
Chapter: Circles
Topic: Introduction to Circles