Update 2: Here is the graph I got for $(x_{1},y_{1})=( 78. 965,12. 354)$, for the parametric circle $(x(t),y(t))$ centered at $(100,100)$
$$x=100+90.135\cos \left( 1.3527+\pi -t\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -t\frac{\pi }{180}\right) .$$
together with the 4 points $(x(t),y(t))$ for $t=0,90,180,270$
$$(x_{1},y_{1})=(x(0),y(0)),(x(90),y(90)),(x(180),y(180)),(x(270),y(270)).$$
You might use the following equations in a for loop with $k=0$ to $k=359$, step $1$:
$$x=100+90.135\cos \left( 1.3527+\pi -k\frac{\pi }{180}\right) ,$$
$$y=100+90.135\sin \left( 1.3527+\pi -k\frac{\pi }{180}\right) .$$
to draw the "orbit" with a 1 degree interval.
Update: corrected coordinates of $(x_{1},y_{1})=(140.5,152)$.
You need to consider the new angle and not only the $1{{}^\circ}$ change. The argument of $\cos$ and $\sin$ is this new angle and not $1{{}^\circ}$.
Let $(x_{c},y_{c})=(160,240)$ be the center of the set of circles and $(x_{1},y_{1})=(140.5,152)$. The radius $r$ is
$$\begin{eqnarray*}
r &=&\sqrt{\left( x_{c}-x_{1}\right) ^{2}+\left( y_{c}-y_{1}\right) ^{2}} \\
&=&\sqrt{\left( 160-140.5\right) ^{2}+\left( 240-152\right) ^{2}} \\
&=&90.135
\end{eqnarray*}$$
Call $(x,y)$ the new coordinates of $(x_{1},y_{1})$ rotated by an angle of $-1{{}^\circ}=-\dfrac{\pi }{180}$ around $(x_{c},y_{c})$ with a radius $r$. The new angle is $\theta'=\theta -\frac{\pi }{180}$, $\theta $ being the initial angle. Then
$$\begin{eqnarray*}
x &=&x_{c}+r\cos \left( \theta -\frac{\pi }{180}\right), \\
y &=&y_{c}+r\sin \left( \theta -\frac{\pi }{180}\right),
\end{eqnarray*}$$
where $\theta $ is the angle $\theta =\arctan \dfrac{y_{1}-y_{c}}{x_{1}-x_{c}}:$
$$\begin{eqnarray*}
\theta &=&\arctan \frac{152-240}{140.5-160}=1.3527+\pi \text{ rad.}\\
&=&\frac{1.3527\times 180{{}^\circ}}{\pi }+180{{}^\circ}=257. 5{{}^\circ}\end{eqnarray*}$$
Thus
$$\begin{eqnarray*}
x &=&160+90.135\cos \left( 1.3527+\pi -\frac{\pi }{180}\right)=
138. 96 \\y &=&240+90.135\sin \left( 1.3527+\pi -\frac{\pi }{180}\right) = 152. 35
\end{eqnarray*}$$
The tangents do not bound the retrograde motion, but the range is close. You are right they define the range for an unmoving outer planet. The boundaries of retrograde motion are found when the velocity of each planet projected on the perpendicular to the line joining them is the same. For a first approximation, you can consider the velocity of the outer planet to be perpendicular to the joining line (valid if the outer planet is much farther from the sun than Earth.) In this approximation, if $\theta$ is the angle at the Earth from the sun vector to the Earth vector is $v_P=v_E \cos \theta$ This supports the tangent for an unmoving outer planet-if we set $v_P=0$ we want $\cos \theta =0,$ so $\theta = \frac \pi 2.$ The boundary of retrograde motion for a moving outer planet will be beyond the tangent when Earth is leading, and behind it when Earth is lagging, but the range will be about the same.
Added: the following assumes circular orbits. Draw the triangle from the planet to the earth and sun. Let $\theta$ be the angle at the earth and $\phi$ the angle at the planet. The projection of the planet velocity perpendicular to the planet-earth line is $v_P \cos \phi$ The projection of the earth velocity perpendicular to the planet-earth line is $v_E \cos \theta$ At the boundaries of the retrograde motion these are equal, so $v_P \cos \phi=v_E \cos \theta$ $\phi$ will very close to the half-width of the Earth's orbit as seen from the planet, so you can solve for $\theta=\arccos \frac {v_P \cos \phi}{v_E}$. You can iterate the solution (improve $\phi$ from the $\theta$ you find) if you need more accuracy.
Best Answer
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.