Q58 / 90 · JEE Main 2017 (02 Apr) · Mathematics

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

A
$2 (\sqrt{2} + 1)$
B
$2 (\sqrt{2} - 1)$
Correct answer
C
$4 (\sqrt{2} - 1)$
D
$4 (\sqrt{2} + 1)$

Solution

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

And $r = |4 - k|$

$\Rightarrow r = 4 - k = \frac{k}{\sqrt{2}}$

$\Rightarrow 4 = \frac{k + \sqrt{2} k}{\sqrt{2}}$

$\Rightarrow k = \frac{4 \sqrt{2}}{\sqrt{2} + 1}$

$\Rightarrow r = \frac{4 \sqrt{2}}{(\sqrt{2} + 1) \sqrt{2}}$

$\Rightarrow r = \frac{4}{\sqrt{2} + 1}$

$\Rightarrow r = 4 (\sqrt{2} - 1)$

Hence, the correct answer is option: 'C'.

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The radius of a circle, having minimum area, which touches the curve y=4-x2 and the lines, y=x is:

Solution

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