Is the graph of $r^2 = 4$ a circle with radius $2$?

Question

If $r^2 = 4$, taking the square root of both sides will give me $r = 2$, so its graph is a circle with radius $2$. Is this correct? I just wanted to make sure because $r^2$ might imply another graph.

Answer 1

It is the circle of radius 2, and also the circle of radius -2. Fortunately these are the same.

Answer 2

You mean taking the square root of both sides will give you $r=2$.

And that's not actually correct: it gives you "$r = 2$ or $r = -2$".

Of course, that's still a circle with radius 2: the coordinates $(r, \theta)$ and $(-r, \theta + \pi)$ both name the same point. However, you might think of it as tracing out the circle twice rather than once.

(depending on your precise definitions, you might require $r$ to be nonnegative, in which case we can reject the $r=-2$ case, as there are no points that both satisfy $r = -2$ and $r \geq 0$)

Answer 3

Yes, it is...and if you still have some doubts remember polar coordinates:

$$\begin{cases}x=r\cos t\\y=r\sin t\end{cases}\implies \color{red}4=r^2=r^2(\cos^2t+\sin^2t)=\color{red}{x^2+y^2}$$