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JEE Main 20172 April
Exam
JEE Main
Year
2017
Date
2 April
Subject
Mathematics
Chapter
Circle

JEE Main 2017 · 2 April · Mathematics · Circle

The radius of a circle, having minimum area, which touches the...

The radius of a circle, having minimum area, which touches the curve y=4−x2y = 4 - {x}^{2} and the lines, y=∣x∣y = \left|x\right| is:The radius of a circle, having minimum area, which touches the curve y = 4 - x ^ 2 and the lines, y = |x| is:

Options

  1. A2(2+1)2\left(\sqrt{2} + 1\right)
  2. B2(2−1)2\left(\sqrt{2} - 1\right)
  3. C4(2−1)4\left(\sqrt{2} - 1\right)
  4. D4(2+1)4\left(\sqrt{2} + 1\right)

Correct answer

C. 4(2−1)4\left(\sqrt{2} - 1\right)

Step-by-step solution

Length of perpendicular from centre of circle to tangent line r=k2r = \dfrac{k}{\sqrt{2}}

And r=∣4−k∣r = \left|4 - k\right|

The radius of a circle, having minimum area, which touches the curve y = 4 - x ^ 2 and the lines, y = |x| is:

⇒r=4−k=k2⇒ r = 4 - k = \dfrac{k}{\sqrt{2}}

⇒4=k+2k2⇒ 4 = \dfrac{k + \sqrt{2}k}{\sqrt{2}}

⇒k=422+1⇒ k = \dfrac{4\sqrt{2}}{\sqrt{2} + 1}

⇒r=42(2+1)2⇒ r = \dfrac{4\sqrt{2}}{\left(\sqrt{2} + 1\right)\sqrt{2}}

⇒r=42+1⇒ r = \dfrac{4}{\sqrt{2} + 1}

⇒r=4(2−1)⇒ r = 4\left(\sqrt{2} - 1\right)

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