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

Options

A. `8 pi (3 - 2 sqrt(2))`
B. `4 pi (3 + 2 sqrt(2))`
C. `8 pi (2 - 2 sqrt(2))`
D. `4 pi (2 - 2 sqrt(2))`

Answer

A

Explanation

Given parabola `y^(2) = 4x`
So, equation of tangent to parabola (i) at point (1,2) is `2y = 2 (x + 1)`
[`:'` equation of the tangent to the parabola `y^(2) = 4` ax at a point `(x_(1), y_(1))` is given by `yy_(1) = 2a (x + x_(1))]`
`implies y = x + 1` ....(ii)
Now, equation of circle, touch the parabola at point (1,2) is
`(x - 1)^(2) + (y - 2)^(2) + lambda (x - y + 1) = 0`
`implies x^(2) + y^(2) + (lambda - 2) x + (-4 - lambda) y + (5 + lambda) = 0` ....(iii)
Also, Circle (iii) touches the x-axis, so `g^(2) = c`
`implies ((lambda - 2)/(2))^(2) = 5 + lambda`
`implies lambda^(2) - 4 lambda + 4 = 4 lambda + 20`
`implies lambda^(2) - 8 lambda - 16 = 0`
`implies lambda = (8 +- sqrt(64 + 64))/(2)`
`implies lambda = 4 +- sqrt(32) = 4 +- 4 sqrt(2)`
Now, radius of circle is `r = sqrt(g^(2) + f^(2) - c)` ,brgt `implies r = |f|` `[:' g^(2) = c]`
`= [(lambda + 4)/(2)| = (8 + 4 sqrt(2))/(2)` or `(8 - 4 sqrt(2))/(2)`
For least area `r = (8 - 4 sqrt(2))/(2) = 4 - 2 sqrt(2)` units
So, area `= pi r^(2) =pi (16 + 8 - 16 sqrt(2)) = 8 pi (3 - 2 sqrt(2))` sq units