calculuspolar coordinates
The graph of the polar equation $r=a+a\cos(\theta)$ is a cardioid. The graph of $r=a+b\cos(\theta)$ when $a<b$ is a limaçon.
What does the graph of $r=b+a\cos(\theta); b>a$ look like? What is the reason for these shapes?
When I plug it into a calculator, it looks like an off-center circle. Is this right? What is making it look the way it does?
For (a), it's worth noting that $\theta=\pi/4$, $\theta=-3\pi/4$, and $r=0$ each put a point $(r,\theta)$ on the "line" $\theta=\pi/4$.
For (b), you need only find $-\pi\leq\theta\leq\pi$ such that $\frac12=1-\cos\theta$.
For (c), do you know how to find the tangent line at a point on a polar curve?
For (d), do you know how to find arc lengths?
I'm going to answer your question in a more-general context- that way, you'll hopefully be able to sketch any (reasonable) polar curve.
Some useful points:
Tangents at the pole: these occur when $r=0$ (when the curve is at the origin).
In this case, tangents at the pole will be the values of $\theta$ satisfying $4\sin(3\theta)=0$ (which I'll leave you to solve).
Maxima- these are points which are furthest from the origin- maxima occur when $\frac{dr}{d\theta}=0$ (and $\frac{d^2r}{d\theta^2}<0$).
In this case, the maxima are (some) solutions to $12\cos(3\theta)=0$ (which, again, I'll leave you to solve).
I'll give you this one for your question- the period is $\frac{2\pi}{\color{green}{3}}$.
That means you can split the graph into $\color{green}{3}$ parts, each of which will be (rotationally) symmetric to the others- one third of the work!
Once you've found and plotted maxima and tangents at the pole, it is now simply a matter of connecting the dots.
The final result is this:
What I've left out are the values of $\theta$ for which $r$ is maximum (in blue), and the values of $\theta$ for which $r$ is minimum (red). Try and do this yourself!
Best Answer
No, it cannot be a circle, because if you vary a parameter continuously from $a=b$, you get a curve that departs from the Cardioid progressively.
With $b<a$, the cusp of the Cardioid becomes the double point of the Limaçon. With $b>a$, that cusp disappears smoothly. The curve remains asymmetric.
We can find the implicit equation of these curves from
$$\rho^2=b\rho+ax$$ or
$$(x^2+y^2-ax)^2=b^2(x^2+y^2).$$
This is not the equation of a circle.