JEE Main · 2021 · 16 Mar 2021 · Shift 1 · Circle

Take $A$ as the origin with $AB$ along the $x$-axis and $AD$ along the $y$-axis, so that $B=(1,0)$, $C=(1,1)$ and $C_1$ is $x^2+y^2=1$.
A circle touching the lines $AB$ and $AD$ has centre $(r,r)$ and radius $r$, and it touches $C_1$ from inside, so the distance between the centres is $1-r$:

Let $E=(1-h,0)$, so that $EB=h$. The line through $C(1,1)$ and $E$ is $x-hy-(1-h)=0$, and it touches $C_2$ when the distance of $(r,r)$ from it equals $r$:

Since $1-r=2-\sqrt{2}=\sqrt{2}\left(\sqrt{2}-1\right)$ and $r=\sqrt{2}-1$, this reduces to $\sqrt{2}(1-h)=\sqrt{1+h^2}$, that is $h^2-4h+1=0$, whose root below $1$ is $h=2-\sqrt{3}$.
Thus $EB=2-\sqrt{3}$, which is $\alpha +\sqrt{3}\beta$ with $\alpha =2$ and $\beta =-1$, so $\alpha +\beta =1$.