Circumference parametrization Let $C=\{(x,y)\in \Bbb R^2: (x-x_0)^2+(y-y_0)^2=r^2\}$ and let $\varphi :[0,2\pi]\to \mathbb{R}^2$, $\theta \mapsto (x_o+r\cos \theta, y_0+r\sin \theta)$, with $r>0$.
I'm trying to prove that $C\subseteq \varphi ([0,2\pi])$.
What I tried:
Let $(x,y)\in C$. Then $$\frac{y-y_0}{r}=\pm \sqrt {1-\left(\frac{x-x_0}{r}\right)^2}$$ and 
$$-1\leq \frac{x-x_0}{r}\leq 1$$
So I set $$\theta=\arccos \frac{x-x_0}{r}$$
There are no issues with the first component of $\varphi (\theta)$.
The second component gives: $$y_0+r\sin \theta=y_0+r\cdot \left(\pm \sqrt{1-(\cos \theta)^2}\right)=y_0+r\cdot\left(\pm \sqrt{1-\left(\frac{x-x_0}{r}\right)^2}\right)$$
Now I'd like to replace $\pm \sqrt{1-\left(\frac{x-x_0}{r}\right)^2}$ with $\displaystyle \frac{y-y_0}{r}$, but I can't due to the possibility of different signs. I even say that they can have different signs. For instance set $x_0=3, y_0=5, x=2, y=5-\sqrt 3$. Then $\displaystyle \frac{y-y_0}{r}<0$, but $\displaystyle \sin \arccos \frac{x-x_0}{r}>0$.
Is there a way to fix what I tried?
If not, what $\theta$ should I take?
Please read $\textbf{carefully}$: I'm not interested in geometric proofs. I want an analytical proof in the fashion of what I tried.
Thank you for your time.
Can anyone help me regarding the comment on Abel's answer? Thanks.
 A: Hint: If your sign comes out wrong, try picking $\phi = 2\pi-\theta$ and note that $\cos(\phi)=\cos(\theta)$, whereas $\sin(\phi) = -\sin(\theta)$.
A: Let a point $(x,y)\in C$ be given. Then at least one of the following holds:
$$({\rm i}) \ x-x_0>0,\quad ({\rm ii}) \ y-y_0>0,\quad ({\rm iii}) \ x-x_0<0,\quad ({\rm iv}) \ y-y_0<0\ .$$
These cases are not disjoint, but in each case a $\theta$ will be produced such that $(x,y)=\phi(\theta)$. Consider the two following examples: The point $(1,1)$ satisfies (i) and (ii), and a careful analysis would show that the two $\theta$ values obtained for this point coincide. On the other hand, the point $(1,-1)$ satisfies (i) and (iv), but the two $\theta$ values obtained for this point differ by $2\pi$.
In case (i) let 
$$\theta:=\arcsin{y-y_0\over r}\  \in\bigl]-{\pi\over2},{\pi\over 2}\bigr[\quad .$$
Then $y=y_0+r\sin\theta$, and $$(x-x_0)^2 =r^2-(y-y_0)^2= r^2(1-\sin^2\theta)=r^2\cos^2\theta\ .$$
As $x-x_0$,  $\cos\theta$, and $r$ are all $>0$ it follows that $x-x_0=r\cos\theta$, or $x=x_0+r\cos\theta$. This proves $(x,y)=\phi(\theta)$ for a certain $\theta\in\ \bigl]-{\pi\over2},{\pi\over 2}\bigr[\ $.
In case (ii) let
$$\theta':=-\arcsin{x-x_0\over r}\ \in\bigl]-{\pi\over2},{\pi\over 2}\bigr[\quad .$$
Then $x=x_0-r\sin\theta'$, and $$(y-y_0)^2 =r^2-(x-x_0)^2= r^2(1-\sin^2\theta')=r^2\cos^2\theta'\ .$$
As $y-y_0$,  $\cos\theta'$, and $r$ are all $>0$ it follows that $y-y_0=r\cos\theta'$, or $y=y_0+r\cos\theta'$. If we now let $\theta:=\theta'+{\pi\over 2}$ then  $(x,y)=\phi(\theta)$ for a certain $\theta \in\ \bigl]0,\pi[\ $.
In case (iii) let 
$$\theta':=-\arcsin{y-y_0\over r}\  \in\ \bigl]-{\pi\over2},{\pi\over 2}\bigr[\quad .$$
Then $y=y_0-r\sin\theta'$, and $$(x-x_0)^2 =r^2-(y-y_0)^2= r^2(1-\sin^2\theta')=r^2\cos^2\theta'\ .$$
As $x-x_0<0$, and  $\cos\theta'$,  $r$ are  $>0$ it follows that $x-x_0=-r\cos\theta'$, or $x=x_0-r\cos\theta'$. If we now let $\theta:=\theta'+\pi$ then  $(x,y)=\phi(\theta)$ for a certain  $\theta\in\ \bigl]{\pi\over2},{3\pi\over 2}\bigr[\ $.
The case (iv) now may be safely left to you.
