JEE Main 2022 — Circle Question with Solution

Since the circle with centre $\left(\alpha ,\beta \right)$ touches the circle $x^2+{\left(y-1\right)}^2=1$ externally and also touches $x-\text{axis}$,

therefore the distance between the centres will be equal to the sum of the radius of both circles

i.e. $\sqrt{{\left(\alpha -0\right)}^2+{\left(\beta -1\right)}^2}=\beta +1$

$\Rightarrow {\alpha }^2=4\beta$

Hence the locus of $L$ will be $x^2=4y$

Now, the area bounded by $x^2=4y \& y=4$ will be

$A=2{\int }_0^4\left(4-\frac{x^2}{4}\right)dx$

$=2{\left(4x-\frac{x^3}{12}\right)}_0^4=\frac{64}{3}$