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 $R\text{, }x \text{ and }\theta$.
On
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.
Here is a diagram of the relevant parts of your question.
The calculations of the additional angles should be obvious. Using the Law of Sines,
$$\frac{\sin(\theta-\phi)}{x}=\frac{\sin(\pi-\theta)}{R}$$ $$\sin(\theta-\phi)=\frac xR\sin(\pi-\theta)$$ $$\theta-\phi=\sin^{-1}\left(\frac xR\sin(\pi-\theta)\right)$$ $$\phi=\theta-\sin^{-1}\left(\frac xR\sin(\pi-\theta)\right)$$ $$\phi=\theta-\sin^{-1}\left(\frac xR\sin\theta\right)$$
That conversion between the second and third lines, from sine to arcsine, is valid since the angle $\theta-\phi$ must be acute.