Since, $x \cos \phi+y \sin \phi=P$ subtends a right angle at the centre $(0,0)$ of hyperbola
$$
\Rightarrow \quad \begin{aligned}
4 x^2-y^2 & =4 a^2 \\
\frac{x^2}{a^2}-\frac{y^2}{4 a^2} & =1
\end{aligned}
$$

Therefore, making the hyperbola equation homogeneous with help of $x \cos \phi+y \sin \phi=P$, we get
$$
\begin{aligned}
& \frac{x^2}{a^2}-\frac{y^2}{4 a^2}=\left(\frac{x \cos \phi+y \sin \phi}{P}\right)^2 \\
& \begin{aligned}
\Rightarrow & x^2\left(\frac{1}{a^2}-\frac{\cos ^2 \phi}{P^2}\right)+y^2\left(\frac{-1}{4 a^2}-\frac{\sin ^2 \phi}{p^2}\right) \\
& -\frac{2 x y \cos \phi \sin \phi}{p^2}=0
\end{aligned}
\end{aligned}
$$
$\therefore$ Coefficients of $x^2+$ coefficients of $y^2=0$
$$
\begin{aligned}
& \Rightarrow \frac{1}{a^2}-\frac{\cos ^2 \phi}{p^2}-\frac{1}{4 a^2}-\frac{\sin ^2 \phi}{p^2}=0 \\
& \Rightarrow \frac{1}{a^2}-\frac{1}{4 a^2}-\left(\frac{\cos ^2 \phi+\sin ^2 \phi}{p^2}\right)=0 \\
& \Rightarrow\left(\frac{4-1}{4 a^2}\right)-\frac{1}{p^2}=0 \quad \Rightarrow \quad \frac{3}{4 a^2}=\frac{1}{p^2} \\
& \Rightarrow P^2=\frac{4 a^2}{3} \quad \Rightarrow \quad p=\frac{2}{\sqrt{3}} a
\end{aligned}
$$

Since, $P$ is also length of perpendicular from $(0,0)$ to line.
$\therefore$ Radius of circle $=P=\frac{2 a}{\sqrt{3}}$.