The area (in sq. units) of the smaller of the two circles that...

The equation of the tangent to the parabola ${y}^{2} = 4ax$ at \left({x}_{1}, {y}_{1}\right)$is$y{y}_{1} = 2a\left(x + {x}_{1}\right) .$</p><p>Equation of tangent to the parabola${y}^{2} = 4x$at \left(1, 2\right)$ is
$y ⋅ 2 = 2\left(x + 1\right)$

$⇒ x - y + 1 = 0 . . . \left(i\right)$

The equation of the family of circles touching a line $l = 0$ at \left({x}_{1}, {y}_{1}\right)$is${\left(x - {x}_{1}\right)}^{2} + {\left(y - {y}_{1}\right)}^{2} + \lambdal = 0 .$</p><p>Equation of family of circle touching the parabola at \left(1, 2\right)$ is

${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} + \lambda\left(x - y + 1\right) = 0 . . . \left(ii\right)$

Since circle touches $x$-axis, hence by putting $y = 0,$ we get

${x}^{2} + \left(\lambda - 2\right)x + \left(\lambda + 5\right) = 0$

As the circle touches the $x$-axis, hence this equation will have only one root.

And we know that a quadratic equation$a{x}^{2} + bx + x = 0, a \ne 0,$ has only one root when its discriminant $D = {b}^{2} - 4ac$ is zero.

Now $D = 0$

$⇒ {\left(\lambda - 2\right)}^{2} - 4 ⋅ 1 ⋅ \left(\lambda + 5\right) = 0$

$⇒ {\lambda}^{2} - 4\lambda + 4 - 4\lambda - 20 = 0$

$⇒ {\lambda}^{2} - 8\lambda - 16 = 0$

Now, applying the Sridharacharya formula, we get

$\lambda = \dfrac{- \left(- 8\right) \pm \sqrt{{\left(- 8\right)}^{2} - 4 \times 1 \times \left(- 16\right)}}{2 \times 1}$

$⇒ \lambda = 4 \pm 4\sqrt{2}$ for smaller circle $\lambda = 4 - 4\sqrt{2} .$

Hence, the equation of the circle is ${\left(x - 1\right)}^{2} + {\left(y - 2\right)}^{2} + \left(4 - 4\sqrt{2}\right)\left(x - y + 1\right) = 0$

$⇒ {x}^{2} + {y}^{2} - 2x\left(1 - 2\sqrt{2}\right) - 2y\left(4 - 2\sqrt{2}\right) + \left(9 - 4\sqrt{2}\right) = 0$

Since, the circle touches the $x$-axis, hence the absolute value of the $y$-co-ordinate of the circle is the radius of the circle.

Thus, the radius of smaller circle $r = 4 - 2\sqrt{2}$ units.

Area of circle $\pi{r}^{2} = \pi{\left(4 - 2\sqrt{2}\right)}^{2}$

$= \pi\left(16 - 16\sqrt{2} + 8\right) = \pi\left(24 - 16\sqrt{2}\right)$

$= 8\pi\left(3 - 2\sqrt{2}\right)$ sq units.