I want to find the intersections of pairs of curves in polar coodinates. As an example, I have three circles with different offsets in a plane which you can see here.
The offsets are: $\exp\left({\frac{ i 2\pi}{3}}\right) | i \in \{0,1,2\} $
which makes the equation for the curves
$\{x,y\}=\{\cos(\theta)+\ \Re(\exp\left(\frac{ i 2\pi n}{3}\right)),\sin(\theta)+\ \Im(\exp\left(\frac{ i 2\pi n}{3}\right))\}$
Now, pairing the curves together and solving for $\theta$ does not give me an answer which I believe is because the angle offset is not the same in the different curves. I believe I should replace $\theta$ in one with $\theta+n \pi$ and solve for n and $\pi$ but I am not getting the correct answers. What is the systematic way of doing this?
update:
The parametric equations are:
$\left( \begin{array}{c} l_1 \\ l_2 \\ l_3 \\ \end{array} \right)=\left( \begin{array}{cc} \sin (u)+1 & \cos (u) \\ \sin (u)-\frac{1}{2} & \cos (u)+\frac{\sqrt{3}}{2} \\ \sin (u)-\frac{1}{2} & \cos (u)-\frac{\sqrt{3}}{2} \\ \end{array} \right)$
Pairing the first two, I have:
$sin (u)+1 =sin (u+2n\pi)-\frac{1}{2}$
$\cos (u) = \cos (u+2n\pi)+\frac{\sqrt{3}}{2} $
but $sin (u+2n\pi)=sin(u)$
so for the first equation, we have:
$sin (u)+1 =sin (u)-\frac{1}{2}$
which does not have any answers?
update2: My above approach is wrong. In both cases, when we shift the origin, we are effectively changing both r and theta. Solving for the original theta is meaningless. it is the transformed radius of the two curves that are equal.
This has nothing to do with polar coordinates. You are given two curves through parametrical representations: $$\eqalign{&\gamma_1:\quad u\mapsto\bigl(x_1(u),y_1(u)\bigr)\quad (a\leq u\leq b),\cr &\gamma_2:\quad v\mapsto\bigl(x_2(v),y_2(v)\bigr)\quad (c\leq v\leq d),\cr}\tag{1}$$ and you want to find their points of intersection. Such a point is generated whenever there is a "time" $u$ for the first curve, and a "time" $v$ for the second such that $\gamma_1(u)=\gamma_2(v)$. Therefore we find these points by solving the system $$x_1(u)=x_2(v),\quad y_1(u)=y_2(v)$$ for $u$ and $v$ and plugging the obtained values into $(1)$.
In the simple example of your figure there are of course geometric considerations that lead to the points of intersection with much less work.