The radius of circle,having minimum area which touches the curve y=4-x² and the lines, y=|x| is :- [Main 2017]

Correct Answer is : 4(2−1)4({2}-1)4(2−1)

Correct Answer is : 4(2−1)4({2}-1)4(2−1)

Test Index

Circles Part 1

Section: Mathematics

Question : 68 of 100

Marks: +1, -0

y=|x|

Solution:

Bonus or 4

x^2+(y-\beta)^2=r^2

x-y=0

\left|\frac{0-\beta}{\sqrt{2}}\right|=r\Rightarrow\beta=r\sqrt{12}

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\pm2\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

∴ co-ordinates of centre

=(0,4-r)

∵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=4(\sqrt{2}-1)

which is given in option 4.

Go to Question:

Prev Question

Next Question