How to express $\phi$ in terms of $R\text{, }x\text{ and }\theta$ Let $S$ be a circle with radius $R$ and center at $O$. 
Let $P$ be any arbitrary point inside circle such that its distance from $O$ is $x$ and the ray $\overrightarrow{OP}$ cuts the circle $S$ at $M$.
Let $N$ be any other point on the circle such that $\angle NPM = \theta$ and $\angle NOM = \phi$.
How can we express $\phi$ in terms of $R\text{, }x \text{ and }\theta$.

First, it's important to realise that we can choose a coordinate system such that $P$ and $M$ are on the horizontal axis passing through the centre $O$ (see figure below). This can be done without any loss of generality.

With respect to the centre $O$, then, the coordinates of the various points of interest are:
$$
P = (x, 0)
\,,\qquad
M = (r, 0)
\,,\qquad
N = (r\cos\phi, r\sin\phi)
$$
But the coordinates of $N$ can be expressed in another way as well, namely,
$$
N = (x + \overline{PN}\cos\theta, \overline{PN}\sin\theta)
$$
Therefore,
$$
\overline{PN}\cos\theta = r\cos\phi - x
\qquad\mbox{and}\qquad
\overline{PN}\sin\theta = r\sin\phi
$$
Eliminating $\overline{PN}$ by dividing the second equation by the first results in
$$
\tan\theta = \frac{r\sin\phi}{r\cos\phi - x}
$$
which, after dividing by $\cos\phi$, can be rewritten as
$$
r\tan\theta - x\tan\theta\sec\phi = r\tan\phi
\qquad(1)
$$
Now we need an equation relating $\sec\phi$ with $\tan\phi$:
$$
\tan^2\phi = \sec^2\phi - 1
\qquad(2)
$$
So, squaring $(1)$,
$$
r^2\tan^2\theta - 2rx\tan^2\theta\sec\phi + x^2\tan^2\theta\sec^2\phi = r^2\tan^2\phi
$$
Using $(2)$, we have
$$
r^2\tan^2\theta - 2rx\tan^2\theta\sec\phi + x^2\tan^2\theta\sec^2\phi = r^2\,(\sec^2\phi - 1) = r^2\sec^2\phi - r^2
$$
which, finally, reduces to a quadratic equation for $\sec\phi$:
$$
(x^2\tan^2\theta - r^2)\,\sec^2\phi - 2rx\tan^2\theta\,\sec\phi + r^2\,(1 + \tan^2\theta) = 0
$$
whose solution is:
$$
\sec\phi = \frac{rx\,\tan^2\theta \pm r\sqrt{(x\tan^2\theta)^2 - (x^2\tan^2\theta - r^2)\,(1 + \tan^2\theta)}}{(x^2\tan^2\theta - r^2)}
$$
The quantity inside the radical simplifies to:
$$
(x\tan^2\theta)^2 - (x^2\tan^2\theta - r^2)\,(1 + \tan^2\theta) =
r^2 + (r^2 - x^2)\tan^2\theta
$$
and we find
$$
\sec\phi = \frac{rx\,\tan^2\theta \pm r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}}{(x^2\tan^2\theta - r^2)}
$$
Since $\cos\phi = 1/\sec\phi$,
$$
\cos\phi = \frac{x^2\tan^2\theta - r^2}{rx\,\tan^2\theta \pm r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}}
$$
To fix the sign, consider that $\theta = 0 \Rightarrow \phi = 0$, so
$$
\cos\phi = \frac{x^2\tan^2\theta - r^2}{rx\,\tan^2\theta - r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}}
$$
Since $x \le r$, the value inside the radical is non-negative and shouldn't cause any trouble. The only possible source of trouble is $\theta = \pi/2$, for which $\tan\theta$ is infinitely large. However, we can see that the solution above can also be written as
$$
\cos\phi = \frac{x^2 - r^2/\tan^2\theta}{rx - r\sqrt{r^2/\tan^4\theta + (r^2 - x^2)/\tan^2\theta}}
$$
Thus, in the limit $\theta \to \pi/2$,
$$
\cos\phi = \frac{x^2}{rx} = \frac{x}{r}
$$
which is the result we'd expect by looking at the figure.
Therefore the final solution is:

$$
\cos\phi = \frac{x}{r} \qquad\mbox{if } \theta = \frac{\pi}{2}
$$
  $$
\cos\phi = \frac{x^2\tan^2\theta - r^2}{rx\,\tan^2\theta - r\sqrt{r^2 + (r^2 - x^2)\tan^2\theta}} \qquad\mbox{if } \theta \ne \frac{\pi}{2}
$$

The solution for the region below the horizontal axis can be obtained from the above, by symmetry.

Edit: Before I start getting down-votes, let me show that my answer agrees with Rory's answer. His answer is:
$$
\phi=\theta-\sin^{-1}\left(\frac xR\,\sin\theta\right)
$$
or
$$
\frac{x}{R}\,\sin\theta = \sin(\theta-\phi) = \sin\theta\cos\phi - \cos\theta\sin\phi
$$
But this, upon division by $\cos\theta\cos\phi$ and multiplication by $R$, is exactly my equation $(1)$ above so the two solutions are equivalent. Mine just happens not to be as elegant as Rory's.
