How to analyze that $\{(x,\phi(x)):x\in[-\varepsilon,\varepsilon]\}$ intersects $\{(x,y):x^2+y^2=\varepsilon^2\}$ at two points? 
Let $\phi \in C^1([-a,a])$ with $\phi(0)=\phi'(0)=0$. Prove that there exist $0<\epsilon<a$ such that $\{(x,\phi(x)):x\in[-\varepsilon,\varepsilon]\}$ intersects $\{(x,y):x^2+y^2=\varepsilon^2\}$ at two points.

My Attempt
I thought that we need to avoid the situation that $(x,\phi(x)) \cdot (1,\phi'(x))=0$, since under this hypothesis two curves have the same tangent line. But I got stuck on writing a detailed proof. Any hints? Thanks in advance!
 A: I will change a single notation by assuming that 
$$\varphi \in C^1([-A,A]) \tag{0}$$
A key point is to consider separately the two domains $\mathbb{R}^-$ and $\mathbb{R}^+$. 
Let us consider from now on the case of domain $\mathbb{R}^+.$
Conditions $\varphi(0)=\varphi'(0)=0$ implies that :
$$\lim_{x\to 0_+}\dfrac{\varphi(x)}{x}=\lim_{x\to 0_+}\dfrac{\varphi(x)-\varphi(0)}{x-0}=\varphi'(0)=0\tag{1}$$
In particular, a direct consequence of (1) is that there exist an interval $V_1:=(0,\varepsilon_1)$ such that :
$$x \in V_1 \ \implies \ |\varphi(x)|<x\tag{1*}.$$
$r>0$ being given, let us define a continuous function $\psi_r$ by 
$$\psi_r(x):=\varphi(x)^2-(r^2-x^2)$$
As $\psi_r(0)<0$ and $\psi_r(r)\geq 0$, by continuity of $\psi_r$, there exist $a, \ 0< a \leq r$ such that 
$$\psi_r(a)=\varphi(a)^2-(r^2-a^2)=0.$$
Interpretation : the circle centered in $0$ with radius $r$ intersects plane curve $(x,\varphi(x))$ in (at least one) point $(a,\varphi(a))$ with $0<a\leq r$ (keep in mind that $a$ depends on $r$).
Let us assume that there is a second point $b > 0, \ b \neq a$ such that, for the same $r$: 
$$a^2+\varphi(a)^2=b^2+\varphi(b)^2=r^2\tag{2}$$
Let us show that we obtain a contradiction for a small enough interval $(0,r)$.
The first equality of (2) is equivalent to :
$$\varphi(a)^2-\varphi(b)^2=b^2-a^2\tag{3}$$
As a consequence :
$$\underbrace{\left|\dfrac{\varphi(a)+\varphi(b)}{a+b}\right|}_{Q_1}\underbrace{\left|\dfrac{\varphi(a)-\varphi(b)}{a-b}\right|}_{Q_2} \ = \ 1\tag{4}$$
Let us now prove that (4) cannot hold if $r$ is in a sufficiently small interval $(0,\varepsilon)$.
Using (1) and Mean Value Theorem, there is an interval $V_2:=(0,\varepsilon_2)$ such that $r \in V_2$ implies $a,b \in V_2$ implying itself $Q_2<1.$
Therefore, using (1*), we obtain as well $|Q_1|<1$ for $a,b \in V_1$.
Defining now :
$V_3:=(0,\varepsilon_3)$ with $\varepsilon_3:=\min(\varepsilon_1,\varepsilon_2)$ :  
it suffices to take any $r \in V_3$. As $a$ and $b$ are $\leq r$, due to (2), (4) cannot hold (the LHS is a product of 2 numbers $<1$). Here is the contradiction.
What has been done for positive axis can be done for negative axis, and we have finished (a single intersection with $x>0$ and a single intersection with $x<0$). 
