Is $r=2\cos(\theta)$ a one-petal polar function?

Question

I'm currently learning about polar functions and their graphs in precalculus, and one of the questions on my homework is to identify the shape of the function $r=2\cos(\theta)$. We were taught that functions in the form $r=a\cos(n\theta)$ where $a>0$ are roses. So, were $a=2$ and $n=1$, wouldn't it form this function and be rose?

I ask because the graph is a perfect circle centered at polar coordinates (1,0).

Answer 1

I'd say that it's convenient to include it in the set of rose curves—it can be useful to think about $r=a\sin n\theta$ for noninteger values of $n$ (even starting at $0$), rather than just integers greater than $1$:

Answer 2

Note that in polar coords

$x = r \cos \theta$

$y= r \sin \theta$

You have that $r = 2 \cos \theta$

That is

Then

$$r^2 = 2r \cos \theta$$

$$x^2+y^2 = 2x$$

$$x^2-2x+1+y^2=1$$

$$(x-1)^2+y^2=1$$

That is a circle centered at $(1,0)$ and radius $1$.