question_answer

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

A)  \[4\left( \sqrt{2}+1 \right)\]                      

B)  \[2\left( \sqrt{2}+1 \right)\]

C)  \[2\left( \sqrt{2}-1 \right)\]                       

D)  \[4\left( \sqrt{2}-1 \right)\]

Correct Answer: D

Solution :

(Bonus or d) \[{{x}^{2}}+{{(y-\beta )}^{2}}={{r}^{2}}\] \[x-y=0\] \[\left| \frac{0-\beta }{\sqrt{2}} \right|=r\Rightarrow \beta =r\sqrt{2}\] \[{{x}^{2}}+{{(y-\beta )}^{2}}=\frac{{{\beta }^{2}}}{2}\] \[\Rightarrow \]\[4-y+{{(y-\beta )}^{2}}=\frac{{{\beta }^{2}}}{2}\] \[\Rightarrow \]\[{{y}^{2}}-y(2\beta +1)+\frac{{{\beta }^{2}}}{2}+4=0\] \[\Rightarrow \]\[{{(2\beta +1)}^{2}}-4\left( \frac{{{\beta }^{2}}}{2}+4 \right)=0\]\[4{{\beta }^{2}}+4\beta +1-2{{\beta }^{2}}-16=0\] \[\Rightarrow \]\[2{{\beta }^{2}}+4\beta -15=0\]              \[\beta =\frac{-4\pm \sqrt{16+120}}{4}=\frac{-4\pm 2\sqrt{34}}{4}\] \[=\frac{-2\pm \sqrt{34}}{2}\Rightarrow \frac{\sqrt{34}-2}{2}\] \[r=\frac{\sqrt{34}-2}{2\sqrt{2}}\] 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 \[\therefore \]\[\left| \frac{0-(4-r)}{\sqrt{2}} \right|=r\,\,\Rightarrow r-4=\pm \,r\sqrt{2}\] \[\therefore \]  \[r=\frac{4}{\sqrt{2}+1}=4(\sqrt{2}-1)\] Which is given in option d.