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The radius of a circle, having minimum a...

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

A

(a) `4(sqrt(2)+1)`

B

(b) `2(sqrt(2)+1)`

C

(c) `2(sqrt(2)-1)`

D

(d) `4(sqrt(2)-1)`

Text Solution

Verified by Experts

The correct Answer is:
D
None
Let the center be `(0,beta)`.
Length of perpendicular from `(0,beta)` to y=x,
`i.e.," "r=|(beta)/(sqrt(2))|`.
So, equation of circle is `x^(2)+(y-beta)^(2)=(beta^(2))/(2)`.

Solving circle and parabola, we get
`4-y+y^(2)-2betay+(beta^(2))/(2)=0`
`ory^(2)-(2beta+1)y((beta^(2))/(2)+4)=0`
Since circle touches the parabola, above equation has equal roots.
`:." "D=0`
`rArr(2beta+1)^(2)-4((beta^(2))/(2)+4)=4`
`rArr2beta^(2)+4beta-15=0`
`rArrbeta=(-2+sqrt(34))/(2)`
Therefore radius `=(beta)/(sqrt(2))=(-2+sqrt(34))/(2sqrt(2))`
None of the option is correct.
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