Tracing a non-concentric circle in cylindrical coordinate system

Question

I am trying to trace the outline of the circle, at a distance d and with radius r, with the variables rho and phi_max as defined on the diagram.

I tried to convert both coordinate systems to Cartesian and tried to solve for rho and phi_max with the Cartesian equation for the circle with the same origin. $(x-d)^2 + y^2 = r^2 .$

I converted the coordinates as:

$$ \rho \cos(\phi_{max}(\rho)) = x \\ \rho \sin(\phi_{max}(\rho)) = \pm y $$ ignoring signs because they'll be squared anyway.

I plug these into the circle equation to get:

$$ \rho^2 - 2d\rho\cos(\phi_{max}(\rho)) + (d^2 - r^2) =0 . $$

I can't solve this equation, it appears that there are 2 variables but only 1 equation to solve for them. I could express $\cos(\phi_{max}(\rho))$ in terms of $\rho$, then use the fact that $\rho$ is bounded ( $[d-r,d+r]$) to solve for each value of $\rho$ and $\arccos(\cos(\phi_{max}(\rho)))$, giving me the sets of values.

However, I am not sure if this would be correct, and more importantly I am concerned that I am missing an easier and more rigorous method.

Note: I am not looking for a singular angle at which the line would be tangent to the circle. $\phi_{max}$ is a function of $\rho$, described at every line that connects to a point on the circle. $\phi_{max}$ is not a maximal angle, it is the name of the variable that I did not change in a different context. Please see the additional image. Image for clarification

Answer 1

I think the only thing you're missing is the distinction between the bounding $\varphi_{\text{max}}$ and the dynamical graphing variables $\rho,\varphi$. You presumably want to get an equation for a function $\rho(\varphi)$ (which really has to be a pair of functions) and you have a good starting point,$$ \rho^2-2\rho d\cos\varphi +d^2 -r^2 = 0 $$(you have an unnecessary extra square, and you are unnecessarily only looking for the maximums).

Instead, by completing the square, we get $$ (\rho - d\cos\varphi)^2 - (d\cos \varphi)^2 +d^2 -r^2 = 0,\\ \rho(\varphi) = d\cos\varphi\pm \sqrt{r^2 - (d\sin\varphi)^2} $$the two functions come from $\pm$ and the domain of applicability is clearly where the argument of the square root is nonnegative, $-r/d<\sin\varphi<r/d,$ so $\varphi_{\text{max}}=\sin^{-1}(r/d)$, except that there is no maximum angle if $r>d$.

Answer 2

At the maximal angle $\phi_{max}$ the line "rho" is tangent to the circle. Therefore, by the Pythagorean theorem $$ d^2=r^2+\rho^2. $$ This can easily be solved for $\rho$.