I'm asked to verify that the set of points described by the cartesian equation $$(x-1)^2 + y^2 = 1$$ is equal to the set of points described by the polar equation $$r = 2 \cos{\theta}, \cos{\theta} > 0$$
Clearly the point described by $(0,0)$ under the Cartesian coordinate system is an element of the first set, since $(0-1)^2 + 0^2 = 1$. However, this point corresponds to polar coordinates of the form $(0, \theta)$, for arbitrary $\theta$, but $0 = 2\cos{\theta} \iff \cos{\theta} = 0 \iff \theta=\frac{\pi}{2} + n\pi$; so there are plenty of pairs $(r, \theta)$ that do satisfy the second condition, but there are plenty of pairs that do not. Do I just need to prove that there exists at least one such pair in the second set? How do I think about problems like this formally, and what would a formal proof of this look like?
Let $(x, y)$ such that $(x-1)^2 + y^2 = 1$. The pair $(r, \theta)$ which corresponds to $(x, y)$ satisfies $x = r\cos{\theta}$, $y = r\sin{\theta}$. We must show that such a pair also satisfies $r = 2\cos{\theta}$.
If $r \neq 0$, we have that $1 = (x-1)^2 + y^2 \iff 1 = (r\cos{\theta} – 1)^2 + (r\sin{\theta})^2 \iff 0 = r^2 – 2r\cos{\theta} \iff r = 2\cos{\theta}$.
My problem is with the case of $r=0$. Is the argument below formal enough?
If $r = 0$, then we must have had that $x = r\cos{\theta} = 0$, $y = r\sin{\theta} = 0$. The pair $(0, \frac{\pi}{2})$ in polar coordinates satisfies $r = 2\cos{\theta}$, and so the origin is also contained in the second set.
Best Answer
In $\mathbb{R}^2$ you can formally pass from Cartesian to polar coordinates in this way: $$\begin{cases} x = \rho \cos \theta \\ y = \rho \sin \theta. \end{cases} $$ Of course I need to specify more about $\rho$ and $\theta$, for now $\rho \geq 0$ and $\theta \in [0,2\pi)$. Note that $$\begin{cases} \mathrm{d}x = \cos \theta \,\mathrm{d}\rho -\rho\sin \theta \,\mathrm{d}\theta \\ \mathrm{d}y = \sin \theta \,\mathrm{d}\rho + \rho \cos \theta \,\mathrm{d}\theta, \end{cases}$$ hence \begin{align} \mathrm{d}x \wedge \mathrm{d}y & = (\cos \theta \,\mathrm{d}\rho -\rho\sin \theta \,\mathrm{d}\theta) \wedge (\sin \theta \,\mathrm{d}\rho + \rho \cos \theta \,\mathrm{d}\theta) \\ & = \rho \,\mathrm{d}\rho \wedge \mathrm{d}\theta. \end{align} Therefore the transformation is singular when $\rho = 0$. Also, reversing the equations above you get $$ \begin{cases} \rho^2 = x^2+y^2 \\ \tan \theta = y/x, \end{cases} $$ so when $\theta = \pi/2+k\pi$, where $k$ is an integer, the inverse transformation is not defined. This is why one usually restricts to the conditions $\rho > 0$ and $\theta \neq \pi/2 +k\pi$. This way the coordinate transformation is invertible.
Now using this change of coordinates your equation $(x-1)^2+y^2 = 1 $ becomes \begin{align} (\rho \cos \theta-1)^2+\rho^2\sin^2\theta -1 & = 0 \\ \rho^2-2\rho\cos \theta& = 0 \\ \rho(\rho-2\cos \theta)&=0, \end{align} which implies $\rho = 2\cos \theta$, because $\rho \neq 0$. Actually $\rho > 0$, which implies $\cos \theta > 0$. This is how you go from one set to the other when the transformation is invertible, but $\rho = 0$ is also a solution of $\rho(\rho-2\cos \theta)=0$, thus just study this case on its own. It is trivial, because $\rho = 0$ corresponds to the origin $(x,y) = (0,0)$.