The correct option is
Slope of line $=2 \Rightarrow \tan \theta=2$
$
\begin{aligned}
& C(0, \alpha) \quad \alpha>0 \\
& \alpha+r=5+\sqrt{5}
\end{aligned}
$
Line $y=2 x$ is tangent to the circle
$
\begin{aligned}
& \therefore\left|\frac{0-\alpha}{\sqrt{4+1}}\right|=r \\
& \Rightarrow|-\alpha|=r \sqrt{5} \\
& \Rightarrow \alpha=r \sqrt{5} \quad \text { as } \alpha>0
\end{aligned}
$
From equation (1) $r \sqrt{5}+r=5+\sqrt{5}$
$
\begin{aligned}
& \Rightarrow \quad r(\sqrt{5}+1)=\sqrt{5}(\sqrt{5}+1) \\
& \Rightarrow \quad r=\sqrt{5}
\end{aligned}
$
And $\alpha=r \sqrt{5}=\sqrt{5} \times \sqrt{5}=5$
Centre $C(0,5)$
$
\begin{aligned}
& \mathrm{OC}=5 \quad A_1 C=\sqrt{5} \\
& \therefore \quad O A_1=\sqrt{25-5}=\sqrt{20}=2 \sqrt{5} \\
& \tan \theta=2 \quad \text { (from figure) } \\
& \cos \theta=\frac{1}{\sqrt{5}} \quad \sin \theta=\frac{2}{\sqrt{5}} \\
& A_1\left(0+O A_1 \cos \theta, 0+O A_1 \sin \theta\right) \\
& A_1\left(2 \sqrt{5} \times \frac{1}{\sqrt{5}}, 2 \sqrt{5} \times \frac{2}{\sqrt{5}}\right) \\
& A_1(2,4)
\end{aligned}
$
Let $B_1\left(x_1, y_1\right)$
$
\begin{aligned}
& \therefore \quad \frac{x_1+2}{2}=0 \text { and } \frac{y_1+4}{2}=5 \\
& x_1=-2 \\
& B_1(-2,6) \quad y_1=6 \\
& \alpha=5 \quad r=\sqrt{5} \quad A_1(2,4) \quad B_1(-2,6)
\end{aligned}
$
