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
$$\begin{eqnarray*}
x &=&160+90.135\cos \left( 1.3527+\pi -\frac{\pi }{180}\right)=
138. 96 \\y &=&240+90.135\sin \left( 1.3527+\pi -\frac{\pi }{180}\right) = 152. 35
\end{eqnarray*}$$
Best Answer
break this up into cases:
case1: absolute value of difference in angles is less or equal to 180 degrees. this is the only case you are considering.
case2: absolute value in difference in angles is greater than 180 degrees. If this is the case, then take 360-(difference in angles)
So in psudo-code, it would look something like:
Degree_Change=|current_degree-last_degree| is what you initially calculate. But you have to make sure that it is less than or equal to 180 degrees, so you look at a couple of cases:
so you do not need to modify it
subtracting by 360 gives you an angle less than 180.