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The area (in sq. units) of the smaller o...

The area (in sq. units) of the smaller of the two circles that touch the parabola, `y^(2) = 4x` at the point (1, 2) and the x-axis is:

A

`4 pi (2 - sqrt(2))`

B

`8 pi (2 - sqrt(2))`

C

`8 pi (3 - 2 sqrt(2))`

D

`4 pi (3 + sqrt(2))`

Text Solution

Verified by Experts

The correct Answer is:
C
`{:(y^(2) = 4x),(2yy' = 4),(y' = (2)/(y)):}|{:(-1 = (beta - 2)/(alpha - 1)),(- alpha + 1 = beta 2),(beta = 3 - alpha):}`
Slope of normal `= (-2)/(2) = - 1`
`(alpha - 1)^(2) + (beta - 2)^(2) = beta^(2)`, `alpha^(2) + beta^(2) - 2alpha = 4 beta +5 = beta^(2)`
`alpha^(2) - 2 alpha - 4 beta + 5 = 0`
`alpha^(2) - 2 alpha - 4 beta + 5 = 0`
`alpha^(2) - 2 alpha - 12 + 4 alpha + 5 = 0`
`alpha^(2) + 2 alpha - 7 = 0`
`(alpha + 1)^(2) = 8`, `alpha + 1 = +- 2 sqrt(2)`
`alpha = - 1 - 2 sqrt(2), -1 + 2 sqrt(2)` `:. beta = 3 - (-1 + 2 sqrt(2))`
Radius `beta = 4 - 2 sqrt(2)`
Area `= pi (beta^(2)) = pi (4 - 2 sqrt(2))^(2) = 4 pi (2 - sqrt(2))^(2) = 4 pi (4 + 2 - 4 sqrt(2)) = 8pi (3 - 2 sqrt(2))`
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