For simplicity, let's assume that the angle $\theta$ is restricted to be in $[0,\frac{\pi}{2}]$, meaning that you always look to the right and up. Let $A(a,b)$ be the point you are given (and assume that $a^2+b^2<R^2$) and let $B,C$ be on the circle such that $B$ is on the line horizontal from $A$, so $B(\sqrt{R^2-b^2},b)$ (again assuming that you look to the right when doing the construction) and $C(\sqrt{R^2-x^2},x)$ is such that $\mu(\widehat{BAC})=\theta$.
Now this will get a little ugly with analytic geometry, but such is life.
We want to write the area of $\triangle ABC$ in two ways. This will give $\frac{AB\cdot AC \sin{\theta}}{2}=\frac{AB\cdot(x-b)}{2}$, where we used the sine area law on the LHS and the fact that the height corresponding to the edge $AB$ is just $x-h$ ($x>b$ as we assumed $\theta>0$). Simplifying this means $AC=\frac{x-b}{\sin{\theta}}$.
On the other hand, we can write down the length of $AC$. $AC^2=(\sqrt{R^2-x^2}-a)^2+(x-b)^2$, so we are left to solve the equation
\begin{equation*}
(\sqrt{R^2-x^2}-a)^2+(x-b)^2=\frac{(x-b)^2}{\sin^2{\theta}}
\end{equation*}
or equivalently
\begin{equation*}
(\sqrt{R^2-x^2}-a)^2=\Big(\frac{(x-b)}{\tan{\theta}}\Big)^2.
\end{equation*}
Since we restricted $\theta\in[0,\frac{\pi}{2}]$, we can take square roots without worrying about signs, so we reduce to
\begin{equation*}
\sqrt{R^2-x^2}=a+\frac{x-b}{\tan{\theta}}.
\end{equation*}
Squaring this should lead to degree two equation for $x$, and then we can get the length $AC$ as $\frac{x-b}{\sin{\theta}}$.
And if $\theta=\frac{\pi}{2}$, one immediately sees that $\sqrt{R^2-x^2}=a$ and can solve this for $x$.
