For $f(t)$ to be in the sphere, i'ts sufficient and necessary that $f'(t)$ and $f(t)-a$ are perpendicular (and it already holds for some $t_0)$ 
Let $f:I\to\mathbb{R}^n$ be a differentiable path. Given
   $a\in\mathbb{R}^n$ and $r>0$, for $f(t)$ to be in the sphere of center
   $a$ and radius $r$ it's suficient and necessary that this occurs for
  some $t_0\in I$ and that the velocity vector $f'(t)$ is perpendicular
  to $f(t)-a$ for all $t\in I$

Of course the first thing I tried to do was to enforce the inner product to be $0$:
$$<f(t)-a,f'(t)> \  = 0 \implies <f(t),f'(t)> - <a,f'(t)> \ = 0$$
I now need to suppose that for some value $t_0$, $f(t)$ is inside the sphere, that is, $|f(t_0)-a|<r\implies \sqrt{<f(t_0)-a,f(t_0)-a>} < r \implies <f(t_0)-a,f(t_0)-a> \  < r^2\implies$
$$<f(t_0),f(t_0)-a>-<a,f(t_0)-a> \ < r^2 \implies $$
$$<f(t_0),f(t_0)>-<f(t_0),a>-(<a,f(t_0)>-<a,a>) \ < r^2$$
But I don't know how it helps in proving that $<f(t)-a,f(t)-a> \ < r^2$ for all $t\in I$ (that is, $f(t)$ is always inside thesphere)
 A: First, a quick word on terminology. There are two geometric objects that are interchanged in daily conversation, but have different mathematical meanings. 


*

*A sphere is just the surface / boundary. For example, a sphere in $\mathbb{R}^n$ centered at $\bf a$ with radius $r$ would be $\{ \mathbf{x} \in \mathbb{R}^n ~:~ \| \mathbf{x} - \mathbf{a} \| =r \}$. If you were to draw a circle on a piece of paper, it would only be the line on the boundary, not the entire region filled in.

*A ball is a sphere and the entire region enclosed by that sphere. A ball in $\mathbb{R}^n$ centered at $\bf a $ with radius $r$ would be $\{ \mathbf{x} \in \mathbb{R}^n ~:~ \| \mathbf{x} - \mathbf{a} \| \leq r \}$. Note the change to "$\leq$" instead of "$=$". If you drew a circle and filled it in, this would be a ball. 
Given that you were trying to use $\|\mathbf{f}(t_0) - \mathbf{a} \| < r$, you were actually dealing with the interior of a ball and not a sphere. 
Solution as follows (mouse-over to reveal).
Hint 1:

 Restating the hypotheses, $\|\mathbf{f}(t_0) - \mathbf{a} \| =r$ for some $t_0$ and $\langle \mathbf{f}(t)-\mathbf{a}, ~ \mathbf{f}'(t) \rangle =0 $ for all $t \in I$. 

Hint 2:

 We need that $\|\mathbf{f}(t) - \mathbf{a} \| =r$ holds for all $t\in I$, i.e., $\|\mathbf{f}(t) - \mathbf{a} \|$ is constant. What do you know about derivatives of constant functions? Also, working with the squared length of vectors is often easier $\dots$

Hint 3:

 Recall that $\| \mathbf{v} \|^2 = \langle \mathbf{v}, \mathbf{v} \rangle$ and $\frac{d}{dt} \langle \mathbf{f}(t), \mathbf{g}(t) \rangle = \langle \mathbf{f}'(t), \mathbf{g}(t) \rangle +\langle \mathbf{f}(t), \mathbf{g}'(t) \rangle$

Solution: 

 Using the above observations, \begin{align*}\frac{d}{dt} \| \mathbf{f}(t)-\mathbf{a} \|^2 &= \frac{d}{dt} \langle \mathbf{f}(t)-\mathbf{a}, \mathbf{f}(t)- \mathbf{a} \rangle \\ &= 2 \langle \mathbf{f}'(t), \mathbf{f}(t)-\mathbf{a} \rangle \\ &= 0 \end{align*} 
 where the last equality holds by our hypothesis. Thus $\|\mathbf{f}(t) - \mathbf{a} \|$ is constant as its derivative is zero. Since $\|\mathbf{f}(t_0) - \mathbf{a} \|=r$ for some $t_0 \in I$, we have that $ \|\mathbf{f}(t) - \mathbf{a} \|=r$ for all $t \in I$. Note that this logic is easily reversed to give the "only if" portion of the proposition.

