Mathematics · Circles

JEE Main 2024 — 31 January, Shift 2 — Question 4

Let a variable line passing through the centre of the circle $x^{2}+y^{2}-16 x-4 y=0$ , meet the positive co-ordinate axes at the point A and B . Then the minimum value of $\mathrm{OA}+\mathrm{OB}$ , where O is the origin, is equal to

Option A:
12
Option B:
18
Correct
Option C:
20
Option D:
24

Answer: B

Step-by-step solution

$(y-2)=m(x-8)$ $\Rightarrow$ x-intercept

$\Rightarrow\left(\frac{-2}{\mathrm{~m}}+8\right)$ $\Rightarrow \mathrm{y}$ -intercept

$\Rightarrow(-8 \mathrm{~m}+2)$

$\Rightarrow \mathrm{OA}+\mathrm{OB}=\frac{-2}{\mathrm{~m}}+8-8 \mathrm{~m}+2$

$f^{\prime}(m)=\frac{2}{m^{2}}-8=0$

$\Rightarrow \mathrm{m}^{2}=\frac{1}{4}$

$\Rightarrow \mathrm{m}=\frac{-1}{2}$

$\Rightarrow \mathrm{f}\left(\frac{-1}{2}\right)=18$

$\Rightarrow$ Minimum $=18$

Answer key and solution verified before publishing.

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Exam: JEE Main 2024
Paper: 31 January, Shift 2
Subject: Mathematics
Chapter: Circles
Topic: Introduction to Circles