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JEE Main 202431 JanuaryEvening Shift
Exam
JEE Main
Year
2024
Date
31 January
Shift
Evening Shift
Subject
Mathematics
Chapter
Circle

JEE Main 2024 · 31 January · Evening Shift · Mathematics · Circle

Let a variable line passing through the centre of the circle x ^ 2...

Let a variable line passing through the centre of the circle x2+y2−16x−4y=0{x}^{2} + {y}^{2} - 16x - 4y = 0, meet the positive co-ordinate axes at the point AandBA\text{and}B. Then the minimum value of OA+OBOA + OB, where OO is the origin, is equal to

Options

  1. A1212
  2. B1818
  3. C2020
  4. D2424

Correct answer

B. 1818

Step-by-step solution

Given: A line passing through the centre of circle x2+y2−16x−4y=0{x}^{2} + {y}^{2} - 16x - 4y = 0 which is (8,2)\left(8, 2\right) intersects positive x and  x \text{\text{ and } } y axis at A and  A \text{\text{ and } } B respectively,

So, let the line be xa+yb=1\dfrac{x}{a} + \dfrac{y}{b} = 1

Now, given line passes through the centre,

So, 8a+2b=1\dfrac{8}{a} + \dfrac{2}{b} = 1

Now, using A.M≥H.MA . M \ge H . M we get,

a2+a2+a2+a2+b1+b16≥62a+2a+2a+2a+1b+1b\dfrac{\dfrac{a}{2} + \dfrac{a}{2} + \dfrac{a}{2} + \dfrac{a}{2} + \dfrac{b}{1} + \dfrac{b}{1}}{6} \ge \dfrac{6}{\dfrac{2}{a} + \dfrac{2}{a} + \dfrac{2}{a} + \dfrac{2}{a} + \dfrac{1}{b} + \dfrac{1}{b}}

⇒2a+2b6≥61⇒ \dfrac{2a + 2b}{6} \ge \dfrac{6}{1}

⇒2a+2b≥36⇒ 2a + 2b \ge 36

⇒a+b≥18⇒ a + b \ge 18

⇒OA+OB≥18⇒ OA + OB \ge 18

Hence, the minimum value of OA+OB=18OA + OB = 18

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