Why $(x-\frac{1}{2})^2 + y^2 = \frac{1}{4}$ is equivalent to $r = \cos(t)$ in polar coordinates?

Question

Repeating the question

Why $(x-\frac{1}{2})^2 + y^2 = \frac{1}{4}$ is equivalent to the function $r = \cos(t)$ in polar coordinates?

The author's solution was

Unfortunately, I don't quite get "a bit" of algebra he is talking about.

So far what I did was

$$\begin{align}(x-\frac{1}{2})^2 + y^2 = \frac{1}{4} &\implies x^2 - x + \frac{1}{4} +y^2 = \frac{1}{4} \\ &\implies x^2 - x+y^2 = 0 \\ & \implies \cos (\theta)^2 r^2 - \cos (\theta) r + \sin^2 (\theta) r^2 =0 \\ & \implies r^2 - \cos \theta r = 0 \end{align}$$

The only possibility (that I see) to obtain the desired result is to divide both sides by $r$ and rearrange. But in this case we need to suppose that $r ≠ 0$. Are we allowed to? And why in his solution he switched variable $\theta$ with $t$?

Answer 1

From your last you can go to $r(r-cos \theta)=0$. You can observe that $r=0$ corresponds to $x=0,y=0$ and is a solution for the original equation. Then you can say that you are going to look for other solutions, where $r \neq 0$, so you can divide by it and get $r=\cos \theta.$ I don't know why the author switched to $t$.

Answer 2

You want to express$C=\{(x,y)\big|(x-1/2)^2+y^2=1/4\}$ in polar coordinates. As you said, making $x=r\cos t, y=r\sin t$ gives you $C=\{(x,y)=(r\cos t,r\sin t)|r(r-\cos t)=0\}$. As you pointed out, there is the chance for $r=0$, but this case corresponds only to the point $(x,y)=(0,0)$. So, we technically have $$C=\{(x,y)=(r\cos t,r\sin t)\big|(r-\cos t)=0\}\cup \{(0,0)\}$$ i.e. $$C=\{(x,y)=(r\cos t,r\sin t)\big|r=\cos t\}\cup \{(0,0)\}$$ but note that the point $(0,0)$ is already considered when $r=0=\cos (\frac{\pi}{2})$. SO, this point is "redundant". We conclude $$C=\{(x,y)=(r\cos t,r\sin t)\big|r=\cos t\}$$ So, the curve is given by the equation $r=\cos t$ in polar coordinates.