Q. 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 :
Solution:
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.
