Question: The area (in sq. units) of the smaller of the two circles that touch the parabola, $\mathrm{y}^{2}=4 \mathrm{x}$ at the point $(1,2)$ and the $x$-axis is :-
$4 \pi(2-\sqrt{2})$
$8 \pi(3-2 \sqrt{2})$
$4 \pi(3+\sqrt{2})$
$8 \pi(2-\sqrt{2})$
Correct Option: , 2
Solution:
