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)

$8\pi (3-2\sqrt{2})$

Equation of the circle is

${(x-1)}^2+{(y-2)}^2+\lambda (x-y+1)=0\Rightarrow x^2+y^2+x(\lambda -2)+y(-4-\lambda )+(5+\lambda )=0$

As circle touches x axis then $g^2-c=0$

$\frac{{(\lambda -2)}^2}{4}=(5+\lambda )$

${\lambda }^2+4-4\lambda =20+4\lambda \Rightarrow {\lambda }^2-8\lambda -16=0\Rightarrow \lambda =\frac{8\pm \sqrt{128}}{2}\Rightarrow \lambda =4\pm 4\sqrt{2}$

Radius $=\left|\frac{(-4-\lambda )}{2}\right|$

Put $\lambda$ and get least radius