Let the center of the circle be $(a,b)$. Drag the center to the origin. This procedure drags our given point $(x,y)$ to $(x-a,y-b)$.
For simplicity of notation, let $u=x-a$, $v=y-b$. Now we determine the angle that the positive $x$-axis has to be rotated through (counterclockwise) to hit the line from the origin to $(u,v)$. Call this angle $\theta$.
Then $\theta$ is the angle, say in the interval $(-\pi,\pi]$, whose cosine is $u/r$, and whose sine is $v/r$, where $r$ is the radius of the circle. (This was already known; it also happens to be $\sqrt{u^2+v^2}$.) So from now on we can take $\theta$ as known. But we have to be careful to take the signs of $u$ and $v$ into account when calculating $\theta$.
Let $D$ be the distance travelled. Assume that we are travelling counterclockwise. Then the angle of travel is (in radians) equal to $D/r$. Let $\phi$ be this angle. If we are travelling clockwise, just replace $\phi$ by $-\phi$. So from now on we can take $\phi$ as known.
After the travel, our angle is $\theta+\phi$. This means that we are at the point with coordinates
$$(r\cos(\theta+\phi), \: \: r\sin(\theta+\phi)).$$
Now transform back, by adding $(a,b)$ to the point. We obtain
$$(a+r\cos(\theta+\phi),\:\: b+r\sin(\theta+\phi)).$$
All the components of this formula are known, so we can compute the answer.
Comment: Note that
$$\cos(\theta+\phi)=\cos\theta\cos\phi-\sin\theta\sin\phi.$$
But we know that $r\cos\theta=u$ and $r\sin\theta=v$.
Thus
$$a+r\cos(\theta+\phi)=a+u\cos\phi-v\sin\phi.$$
Similarly,
$$\sin(\theta+\phi)=\cos\theta\sin\phi+\sin\theta\cos\phi,$$
and therefore
$$b+r\sin(\theta+\phi)=u\sin\phi+v\cos\phi.$$
Thus an alternate (and for many purposes simpler) version of the answer is
$$(a+(x-a)\cos\phi-(y-b)\sin\phi,\:\: b+(x-a)\sin\phi+(y-b)\cos\phi),$$
where $\phi=D/r$.
We could also reach this by quoting the rotation formula. Recall that when we are rotating a point $(u,v)$ about the origin through an angle $\phi$, we multiply the vector by a certain matrix. You can think of this post as being, in particular, a derivation of the rotation formula.
If the orbit is circular the calculation is simple: $A_x=S_x+r \cos (ts), A_y=S_y+r \sin (ts)$. Without the force law you don't know if the orbit is circular, which will happen if the speed is just right for the distance. Presumably you are using Newton's gravitational law, in which case you need the mass of the sun to plug in.
If the orbit is not circular, you can either solve Kepler's laws or integrate the differential equation. The differential equation lets you apply other forces that might change the orbit as you go along, like thrusting from a spaceship or gravity effects of other planets. A popular approach for the differential equation is a Runge-Kutta method, where $y$ in the Wikipedia page becomes a vector with one component for each space direction of location and another for each component of velocity.
Best Answer
That's simple: if $r$ is the radius and angles, measured w.r.t. the vertical axis, are in radians, the distance travelled is $d=r\theta$.
On the other hand, $x=r\cos\bigl(\frac\pi2+\theta\bigr)=-r\sin\theta\:$ and $\:y=r\sin\bigl(\frac\pi2+\theta\bigr)-r =r(\cos\theta -1) $, so that \begin{cases} x=-\dfrac{d\sin \theta}\theta,\\[1ex] y=\dfrac{d(\cos\theta-1)}\theta. \end{cases}