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

Answer 1

Answer: (B)

Equation of circle is
${\left(x-1\right)}^2+{\left(y-2\right)}^2+\lambda \left(x-y+1\right)=0$
$\Rightarrow x^2+y^2+x\left(\lambda -2\right)+y\left(-4-\lambda \right)+\left(5+\lambda \right)=0$
As cirlce touches x axis then $g^2-c=0$
$\frac{{\left(\lambda -2\right)}^2}{4}=\left(5+\lambda \right)$
${\lambda }^2+4-4\lambda =20+4\lambda$
${\lambda }^2-8\lambda -16=0$
$\lambda =\frac{8\pm \sqrt{128}}{2}$
$\lambda =4\pm 4\sqrt{2}$
Radius $=\left|\frac{\left(-4-\lambda \right)}{2}\right|$
Put $\lambda$ and get least radius.