Parametrizing a set I'm trying to parameterize the following set
$$
\left\lbrace\, r\left(\cos\left(\alpha\right),\sin\left(\alpha\right)\right)  \;\middle\rvert \quad 0\leq\alpha\leq2\pi, \quad 0<r\leq 1,\quad \tfrac{1}{r}-\alpha=2k\pi\ \,\text{ for some } k\,\right\rbrace
$$
The corresponding curve $\gamma:[a,b]\to\mathbb{R}^2$ will probably look similar to an archimedean spiral, as I've worked with the components separately, i.e. fixing an $\alpha$ or $r$. In general, I know the process of finding a parametric expression, but the last condition, i.e. $1/r-\alpha$ being a multiple of $2\pi$ makes me unsure what to do.
 A: Assuming the allowable $k$ are the integers $1,2,\dots,$ the curve $\gamma:[0,\infty)\to \mathbb R^2$ given by
$$\gamma (t) = \frac{1}{2\pi +t}(\cos t,\sin t)$$
would appear to parameterize your set.
Note that the set $S$ you've described is not compact. So there can be no continuous map from a closed bounded interval onto $S.$ If you want to include $(0,0)$ in the set, which may be natural, then we can change variables and find a reasonable parameterization $\varphi :[0,1] \to S\cup \{(0,0\}.$
A: The last condition says $\alpha = r^{-1} + 2\pi k$ for some integer $k$, which has an infinite number of solutions for $\alpha$, evenly spaced by $2\pi$.  Any such solution generates the same point $r(\sin(\alpha),\cos(\alpha))$, namely $r(\sin(r^{-1}),\cos(r^{-1}))$, and at least one of the solutions satisfies $0\leq \alpha \leq 2\pi$, so this point does lie in the set.  
Hence, for any $r$ you get a unique point $r(\sin(r^{-1}),\cos(r^{-1}))$ in the set.  So your set is just $\{r(\sin(r^{-1}),\cos(r^{-1})) \;| \; 0<r\leq 1\}$. 
A: $$ \dfrac{1}{r} - \alpha = 2 k \pi $$
$$ \sin(\dfrac{1}{r} - \alpha) = 0 $$
Any other trig circular function of period $2\pi$ could be chosen. 
$$ \sin(\dfrac{1}{r} ) \cos \alpha= \cos(\dfrac{1}{r} ) \sin \alpha $$
$$ \tan(1/r) = \tan \alpha $$
$$ 1/r = \alpha, 1/r = \pi+ \alpha $$
and other overlapping or coinciding co-terminals for all integers $k$, the spiral shapes are periodic or self-repeating as belonging  to a polar cyclic infinite set. For each $r$ there are an infinite number of  $\alpha^{'}$s.
For arctan function we have $ x,y,r $ of opposite sign in third quadrant compared to first quadrant.
So we have two solutions. The polar equations 
$$ r\alpha = +1, r\alpha = -1 $$
are the solutions represented by the two hyperbolic spirals shown:

If a separarate parametrization is required for the two curves we simply have
$$ r=1/t, \alpha = +t\, ;\, r=1/t, \alpha = -t\,;$$
The magenta spiral has higher $\alpha$ domain, overlaps blue line due to plotting as a second graph. 
From this solution you can choose any $(\alpha,\, r)$ from the set as desired.
