How to prove that the circumference is $C^k$? Notation:
$U$ is a open subset of $\mathbb{R}^n$;
$\partial U$ is a boundary of $U$;
A $C^k$ function is a function $k$-times continuosly differentiable;
$B(x_0,r)$ is a ball in $\mathbb{R}^n$ with center $x_0$ and radius $r$;
$x=(x_1,...,x_n)$ is a point of $\mathbb{R}^n$.

Definition: We say $\partial U$ is $C^k$ is for each point $x_0\in \partial U$ there exist $r>0$ and a $C^k$ function $\gamma:\mathbb{R}^{n-1}\to \mathbb{R}$ such that - upon relabeling and reorienting the coordinates axis if necessary - we have $$U\cap B(x_0,r)=\{x\in B(x_0,r);\;x_n>\gamma(x_1,...,x_{n-1})\}$$ (Lawrence C. Evans, Partial Differential Equations)

Problem: Let $U=\{x_1^2+x_2^2<1\}$. How to prove that $\partial U$ is $C^k$?
My interpretation of problem (probably wrong) can be summarized in the following picture.

Thanks.
 A: 
To solve the problem, should I find a $C^k$-curve like the red curve below?

No. Well, sorta. Please pardon my initial carelessness. You have to prove: for every point on the boundary, the boundary near this point (every pde book or notes like to use a ball intersecting the domain to characterize this) has no difference with a graph $C^k$-function of which the domain is codimension 1. Or more simply put, locally speaking the boundary is the graph of a $C^k$-function from codimension 1 (upon relabeling and reorienting the coordinates axis). 

Let $U=\{x_1^2+x_2^2<1\}$. How to prove that $\partial U$ is $C^k$?

By the heuristics above, the boundary, which is $\partial U=\{x ^2+y^2=1\}$ (I am using $x$ and $y$ since we are only dealing with 2d), locally speaking, this set can be view as a curve of a smooth function. 
For example, for any point $(a,b)$ in the first quadrant, the boundary curve in a small neighborhood of it is the graph of $y=\sqrt{1-x^2}$, which is a $C^k$-function for any $k$. 
The only problematic points are the ones near $(1,0)$ and $(-1,0)$. If we don't re-orient the axis, $y$ will become double valued ($y=\pm\sqrt{1-x^2}$) and the tangent grows unbounded. 
However, based on the definition Evans used, we can relabel and reorient the coordinates axes, and this can be done locally for each individual point! This means that we can view the curve near $(1,0)$ and $(-1,0)$ locally, what is the curve like, and don't care if there is a global representation of other parts of $\partial U$!

Lastly, let's use Evans' definition to prove it.
Sketch of the proof: In this case $n=2$, and consider the question for a fixed $k$.
For some 
$p_0:=(x_0,y_0)\in \partial U =
\{x ^2+y^2=1\}$, say for $(x_0,y_0)$, the difference of $x$-coordinate $x_0$ to either $(1,0)$ and $(-1,0)$ is strictly greater than some fixed $0<\delta<1$. By the definition, we wanna find a $C^k$-function from $\mathbb{R}^1$ to $\mathbb{R}^1$ so that 
$$U\cap B(p_0,r)=\{(x,y)\in B(p_0,r);\;y>\gamma(x )\}.\tag{1}$$
For the bottom part excluding the neighborhood near $(1,0)$ and $(-1,0)$, $\gamma(x)$ is $\gamma_0 := -\sqrt{1-x^2}$ for $-1+\delta<x<1-\delta$. Take $r$ to the minimum of the distances from $(x_0,y_0)$ to $p_1:=(1-\delta,-\sqrt{1-(1-\delta)^2})$, and $(x_0,y_0)$ to $p_2:=(-1+\delta,-\sqrt{1-(1-\delta)^2})$. Within $U\cap B(p_0,r)$, this $\gamma=\gamma_0 \in C^k$ for any $k$ because we set $x$ away from $1$ (by at least a fixed $\delta$). At $p_1$ and $p_2$, we can compute the derivatives of $\gamma_0$ up to this arbitrary fixed $k$-th order. Now we can use this information to get $\gamma_1$ and $\gamma_2$, which can be two polynomials. For example, we can explicitly construct $\gamma_1$ by Hermite interpolation using the value $\gamma_0(1-\delta),\dots,\gamma_0^{(k)}(1-\delta)$. Now just glue $\gamma_0$ with $\gamma_1$ and $\gamma_2$ together at $p_1$ and $p_2$. $\gamma_1$ and $\gamma_2$ have their function values and all first to $k$-th derivatives agreeing with $\gamma_0$ at $p_1$ and $p_2$ respecively. We have 
$$\gamma(x)=\begin{cases}
\gamma_0 & \text{ for } -1+\delta<x<1-\delta,
\\
\gamma_1 & \text{ for } x\geq 1-\delta,
\\
\gamma_2 & \text{ for } x\leq  -1+\delta.
\end{cases} $$
This $\gamma\in C^k:\mathbb{R}^1\to \mathbb{R}^1$ for predetermined $k$. We can check the definition (1) holds for any $p_0$ with $x$-coordinate in $(-1+\delta,1-\delta)$.
For the top part, flipping the $y$-axis bottom-up, and we end up in the first case. For the left and right neighborhoods near $(1,0)$ and $(-1,0)$, relabel the $x$ axis to be $y$, $y$ axis to be $x$, we end up in the situations of first two cases.

Last remark: For a more mathematically strict definition, you can refer to the Wikipedia's entry of Lipschitz domain. This is the same one with the definition used in Gilbarg and Trudinger, just replace "Lipschitz continuous functions" with $C^k$-, $C^{k,\alpha}$-, or $C^{\infty}$-functions so that you get the domain of $C^k$-, $C^{k,\alpha}$-, or $C^{\infty}$-smoothness.
