Manually sketching a polar graph by logically transferring from $\theta r$ plane to xy plane

Question

It's not a big problem to draw a polar graph, just like any graph, by computing a necessary amount of points, and then linking them.

But I am now trying to decipher a more sophisticated approach suggested by the authors of the Precalculus book. They base their approach, if I have correctly understood them, on logical transformation of the fundamental cycle from $\theta r$ plane to xy-plane.

I will be grateful if you can help me to understand the logic, its steps, and guide me through the process.

The text book

Based on the example number 2 on pages 945 - 947 (the main first explanation is given on the page 941) I tried to solve the equation:

$$r = 3-5\cos(\theta)$$

I did get a nice limacon, but I did so by computing each x and y value for $2π$ interval. But I would like to do so by logically (that is without computations I did) transferring values of from $\theta r$ plane to xy-plane, just like authors of the book did (manually, visually, without any automatic tools, and without scrupulously computing each x and y values).

Here are my computations:

Here is the way the limacon looks - the same one I got on my paper:

limacon

And here how the answer to the equation looks in the book - please, notice that they also use $\theta = \arccos(\frac{\pi}{3})$ and $\theta = 2π -\arccos(\frac{\pi}{3})$ as sort of asymptotes (those are not asymptotes, but these rays do look like asymptotes because the curve barely touches the line and than bounces off of it).

But how shall I logically deduce, first from having drawn the fundamental cycle on $\theta r$ plane as it's done in the book, and then without explicitly computing $x$ and $y$, draw a limacon on $xy$-plane? For example, the first curve on $xy$-plane shall go from $(-2,0)$ to $(0, 3)$ and move on, making a nice round trip till it reaches $(-8,0)$, but how can I deduce that from the graph I have on the $\theta r$ plane? I understand how the angles "move", no issues with those, but I don't see how I can know the length of each ray such that it starts, say, from $(-2,0)$ and gets to $(0, 3)$ without touching the "asymptote" $\theta = \arccos(\frac{\pi}{3})$?

Thank you very much!