Bisect the angle to get 2 right triangles with known hypotenuse and angles, then use $\sin$ to get the sides opposite the $22.5/2$ degree angles.
Or, use the triangle that's already there, having 2 angles equal to $(180-22.5)/2$ degrees, and apply the law of sines.
Or, apply the law of cosines immediately.
In case you want exact forms for the sines or cosines involved, you can use half/double angle formulas and the fact that $22.5=45/2$.
I'm assuming you know the center $\,M\,$ , so we get that:
$$\angle ORM=90^\circ\Longrightarrow r^2=MR^2=OM^2-OR^2$$
applying Pythagoras Theorem.
Added: Let $\,R=(x_0,y_0)\,\,,\,\,S=(x_0,-y_0)\,$ , so that if $\,M=(a,0)\,$ we then have:
$$\begin{align*} x_0^2+y_0^2=&16\\(x_0-a)^2+y_0^2=&r^2\end{align*}$$
From the second equation we get
$$a^2-2ax_0-r^2+16=0\Longrightarrow a=\frac{2x_0\pm\sqrt{4x_0^2+4r^2-64}}{2}=x_0\pm\sqrt{x_0^2+r^2-16}$$
Thus
$$OM=a=x_0+\sqrt{x_0+r^2-16}$$
and inputting in (**) above we get
$$r^2=2x_0^2+r^2-16+2x_0\sqrt{x_0^2+r^2-16}-16\Longrightarrow (32-2x_0^2)^2=4x_0^2(x_0^2+r^2-16)\Longrightarrow$$
$$1,024-128x_0^2+\rlap{\;/}4x_0^4=\rlap{\;/}4x_0^4+4r^2x_0^2-64x_0^2\Longrightarrow r^2=\frac{256-16x_0^2}{x_0^2}=16\frac{16-x_0^2}{x_0^2}=$$
$$=\left(\frac{4y_0}{x_0}\right)^2\Longrightarrow r=4\frac{y_0}{x_0}=4\tan\angle ROM$$
Best Answer
If $X$ is the centre of the circle, and $r$ is its radius, then by Pythagoras in triangle $BCX$, $BX^2=r^2=(r-3)^2+4^2$.
So you just have to solve the equation $$r^2=(r-3)^2+16$$