Question

The area (in s 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)

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

Answer 2

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 $g^{2} -c=0$ $\frac{\left(\lambda-2\right)^{2}}{4} = \left(5+\lambda\right)$ $\lambda^{2} + 4-4\lambda=20 +4\lambda$ $\lambda^{2} -8\lambda-16 =0$ $\lambda = \frac{8\pm \sqrt{128}}{2}$ $\lambda = 4 \pm4\sqrt{2}$ Radius $= \left|\frac{\left(-4-\lambda\right)}{2}\right|$ Put $\lambda$ and get least radius.