geometry - Why, conceptually, do limaçons $r=a+bcostheta$ have dimples when $|frac{a}{b}|

Using calculus, I can justify that limaçons—the polar graphs of $r=a+b\cos\theta$ for various nonzero real values of $a$ and $b$—are dimpled when $|\frac{a}{b}|<2$, but that doesn't seem to yield any conceptual reason why that should be the case. The boundary value 2 seems too nice to not have a conceptual explanation, so is there one?

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The calculus: For this sort of limaçon, with cosine, there are always vertical tangents to the graph at $\theta=k\pi$, and the dimple is characterized by a pair of vertical tangents near one of those two locations, but equally spaced before and after it, whereas a non-dimpled limaçon only has those two vertical tangents. Vertical tangents occur when $\frac{dy}{dx}$ is undefined; for polars, that means when $\frac{dy}{d\theta}$ is undefined (for our limaçon, never) or when $\frac{dx}{d\theta}=0$. For our limaçon, $\frac{dx}{d\theta}=-\sin\theta(a+2b\cos\theta)$, so $\frac{dx}{d\theta}=0$ implies $\sin\theta=0$ ($\theta=k\pi$) or $a+2b\cos\theta=0$. This latter case, which can be rewritten as $\cos\theta=-\frac{a}{2b}$, has no solutions when $|\frac{a}{2b}|>1$, a single solution that is already in the solutions from $\sin\theta=0$ when $|\frac{a}{2b}|=1$ (so, no additional vertical tangents and hence no dimple when $|\frac{a}{b}|\ge 2$), and two solutions when $|\frac{a}{2b}|<1$ (so two additional vertical tangents and hence a dimple when $|\frac{a}{b}|<2$).