### JEE Main Solved Paper 2017 Question 25

##### Question: The radius of a circle, having minimum area, which touches the curve $ y=4x^{2} $ and the lines, $ y=,|x| $ :- [JEE Main Solved Paper-2017]

#### Options:

A) $ 4( \sqrt{2}+1 ) $

B) $ 2( \sqrt{2}+1 ) $

C) $ 2( \sqrt{2}-1 ) $

D) $ 4( \sqrt{2}-1 ) $

## Show Answer

#### Answer:

Correct Answer: D

#### Solution:

(Bonus or d) $ x^{2}+{{(y-\beta )}^{2}}=r^{2} $ $ x-y=0 $ $ | \frac{0-\beta }{\sqrt{2}} |=r\Rightarrow \beta =r\sqrt{2} $

$ x^{2}+{{(y-\beta )}^{2}}=\frac{{{\beta }^{2}}}{2} $

$ \Rightarrow $ $ 4-y+{{(y-\beta )}^{2}}=\frac{{{\beta }^{2}}}{2} $

$ \Rightarrow $ $ y^{2}-y(2\beta +1)+\frac{{{\beta }^{2}}}{2}+4=0 $

$ \Rightarrow $ $ {{(2\beta +1)}^{2}}-4( \frac{{{\beta }^{2}}}{2}+4 )=0 $ $ 4{{\beta }^{2}}+4\beta +1-2{{\beta }^{2}}-16=0 $

$ \Rightarrow $ $ 2{{\beta }^{2}}+4\beta -15=0 $ $ \beta =\frac{-4\pm \sqrt{16+120}}{4}=\frac{-4\pm 2\sqrt{34}}{4} $ $ =\frac{-2\pm \sqrt{34}}{2}\Rightarrow \frac{\sqrt{34}-2}{2} $ $ r=\frac{\sqrt{34}-2}{2\sqrt{2}} $

which is not in options therefore it must be bonus. But according to history of JEE-Mains it seems they had following line of thinking.

Given curves are $ y=4-x^{2} $ and $ y=,|x| $ There are two circles satisfying the given conditions. The circle shown is of least area. Let radius of circle is ‘r’

$ \therefore $ co-ordinates of centre = (0, 4 - r) $ \because $ circle touches the line $ y=x $ in first quadrant

$ \therefore $ $ | \frac{0-(4-r)}{\sqrt{2}} |=r\Rightarrow r-4=\pm ,r\sqrt{2} $

$ \therefore $ $ r=\frac{4}{\sqrt{2}+1}=4(\sqrt{2}-1) $ Which is given in option d.