Question

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. $4\pi (3+\sqrt{2})$
• B. $8\pi (2-\sqrt{2})$
• C. $8\pi (3-2\sqrt{2})$
• D. $4\pi (2-\sqrt{2})$

Answer 1

The correct option is C $8\pi (3-2\sqrt{2})$

Equation of tangent to circle/parabola at $(1,2)$ is :
$y-2=x-1\Rightarrow x-y+1=0$
Therefore, equation of normal through $(1,2)$ will be $x+y-3=0$
Since, the centre lies on the normal.
Let the coordinates of the centre be $(3-r,r)$
Distance between $(1,2)$ and $(3-r,r)$ is $r$
$\therefore (3-r-1{)}^2+(r-2{)}^2=r^2$
$\Rightarrow \sqrt{2}(r-2)=\pm r$
$\Rightarrow r=\frac{2\sqrt{2}}{\sqrt{2}\mp 1}$
or, $r=2\sqrt{2}(\sqrt{2}-1)$
(As $3-r>0$)
Area of circle $=8\pi (3-2\sqrt{2})$ sq. units