Graphing Polar equation

Question

I was plotting graphs on Desmos, I plotted $r=a+sin(\theta)$ and it rightfully spit out a cardiod at $a=-1$ and $a=1$

Then when I was sliding the values of $a$, it spits out a limacon for $-1<a<1$($a\neq0$)

Then I notice a weird thing that is when I was sliding the value between $1<|a|<3$ it was spitting out something between cardiod and a circle. What is it called?

For values $|a|>3$ it was mapping circles. Why does this back and forth happen between circles, limacons, cardiodes and the unknown shape?

Answer 1

The figure is called a limaçon for all $a \neq 0$, in particular it's called a cardioid for $|a|=1$. In general, the polar equation $r = a + b\sin\theta$ produces a limaçon when $a,b \neq 0$, and a cardioid when $|a/b|=1$.

What it doesn't produce is circles. Remembering $r^2 = x^2 + y^2$ and $r\sin\theta = y$, we find $$x^2 + y^2 = r^2 = r(a + \sin \theta) = a \sqrt{x^2 + y^2} + y $$ and thus $$(x^2 + y^2-y)^2 = a^2(x^2 + y^2) $$ which is the Cartesian equation of a quartic curve for all $a \neq 0$ (for $a=0$ it does simplify to a degree 2 equation, specifically that of a circle, but you're excluding that case). It would however describe a proper circle for all $a \neq 0$ if the $-y$ term on the LHS weren't there.

Answer 2

In fact, all of these graphs are limaçons. The generic equation for a limaçon in polar coordinates is $r = a + b \cos (\theta)$ for a limaçon oriented horizontally, along the $x$-axis, or $r = a + b \sin(\theta)$ for a limaçon oriented vertically, along the $y$-axis. There are different types of limaçons, depending on the relationship between $a$ and $b$ in the polar equation.

1. If $a < b$ then you will have a looped limaçon
2. If $a = b$ then you will have a cardioid. A cardioid is actually a type of limaçon, i.e. the only limaçon that has a cusp.
3. If $b < a < 2b$ then you will have a dimpled limaçon. This is the "weird thing" you noticed when you had the slider in the range roughly between $1$ and $3$.
4. If $a \ge 2b$ then you will actually have a convex limaçon. At exactly $a=2b$, the dimple will flatten out, and if $a > 2b$, then it will even start to bow outwards again... however, it is still not a circle! As $a \gg 2b$, the difference between the limaçon and a circle is hardly discernible, but it is there.

You can tell that your limaçon is not a circle, by plotting an actual circle with similar parameters (Desmos link), and (1) noticing the difference between the two equations, and (2) noticing the visual difference on a graph. Move the sliders for $a$ and $b$ to scale the limaçon. See how, for $a \ge 2b$, it looks quite circleish, but remains very slightly "fatter" along the horizontal direction than a true circle.

Comparing the polar-form equations gives more insight into why exactly this happens.

Before we get into that, however, I'll just take a brief detour and show the derivation for the circle equation in the first place. I simply started with the Cartesian equation for a circle of radius $a$ centered at $(0, b)$, and converted to polar coordinates. (EDIT: this derivation is only for the inscribed circle; I have added also the circle circumscribed about the limaçon to the Desmos graph, but I will not show the derivation of that circle here.) \begin{align} x^2 + (y-b)^2 &= a^2 \\ x^2 + y^2 - 2by + b^2 &= a^2 \\ r^2 - 2br\sin(\theta) &= a^2 - b^2 \\ r^2 - 2br\sin(\theta) + b^2\sin^2(\theta) &= a^2 - b^2 + b^2\sin^2(\theta) \\ (r - b\sin(\theta))^2 &= a^2 - b^2(1 - \sin^2(\theta)) \\ r &= b\sin(\theta) + \sqrt{a^2 - b^2\cos^2(\theta)} \end{align}

OK, now that that's out of the way - let $r(\theta)$ be the limaçon, and let $\hat{r}(\theta)$ be the circle (the equation we just found a moment ago), as polar functions of $\theta$. To make the difference between them crystal clear, we can do a little convenient rearranging, and write both equations in terms of the ratio $a/b$, so that as $a/b\to\infty$, the limaçon gets closer to the circle.

First, the limaçon: \begin{align} r(\theta) &= a + b\sin(\theta) \\ &= b\left( \sin(\theta) + \frac{a}{b} \right) \\ \end{align}

And second, the circle: \begin{align} \hat{r}(\theta) &= b\sin(\theta) + \sqrt{a^2 - b^2\cos^2(\theta)} \\ &= b\left( \sin(\theta) + \sqrt{\left(\frac{a}{b}\right)^2 - \cos^2(\theta)} \right) \\ &\approx b\left( \sin(\theta) + \sqrt{\left(\frac{a}{b}\right)^2} \right) \text{ if }a\gg b \\ &\approx b\left( \sin(\theta) + \frac{a}{b} \right) = r(\theta) \end{align}

As $a$ becomes very much larger than $b$, the ratio $a/b\to\infty$. On the other hand $-\cos^2(\theta)$ will never be less than $-1$ or greater than $0$. So for $a\gg b$, this term becomes essentially insignificant; it is "dwarfed" by the enormous $a/b$ term under the square root. This term, which is tiny for $a\gg b$, is where the difference between your almost-circle limaçon & a true circle enters in.