I've solved this problem, but why is this differentiable? Let $\alpha:\mathbb R\rightarrow \mathbb R^3$ be a smooth curve (i.e., $\alpha \in C^\infty(\mathbb R)$). Suppose there exists $X_0$ such that for every normal line to $\alpha$, $X_0$ belongs to it. Show that $\alpha$ is part of a circumference.
Part of solution: Let $n(s)$ be the normal vector to $\alpha$ on $s$. For every $s \in \mathbb R$ there exists an unique real number $\lambda$ such that $\alpha(s)+\lambda n(s)=X_0$.
Now let $\lambda(s)$ be the function that associates every real number $s$ to this number.

I've finished this question, but I have derivated $\lambda(s)$ and I don't know why I can do it. Can someone tell me why is $\lambda$ differentiable?
 A: You can write
$$\lambda(s)=\langle X_0-\alpha(s), n(s)\rangle$$
supposing $\|n(s)\|=1$ (otherwise you divide by this norm).
That expression is obtained with sums and products of smooth functions (at most divisions by non-vanishing smooth functions), hence it is smooth.
A: You can find an explicit expression for $\lambda$. If $x_0 = \alpha + \lambda n$ then $x_0 - \alpha = \lambda n$. "Dotting" both sides with $n$ gives $(x_0-\alpha)\cdot n = \lambda n \cdot n$, where the dot is the dot product, also called the scaler product.
Since $n$ is a unit vector and $n \cdot n = \|n\|^2$ we have $n \cdot n = 1$ and so $\lambda = (x_0 - \alpha) \cdot n$.
We can go further and actually differentiate $\lambda$. Assuming that $\alpha$ is parametrised by arc-length we have
\begin{array}
1\lambda' &=& (x_0-\alpha)'\cdot n + (x_0-\alpha)\cdot n' \\
&=& -t\cdot n + (x_0-\alpha)\cdot (-\kappa t) \\
&=& -\kappa (x_0 - \alpha) \cdot t
\end{array}
where $t$ is the unit tangent vector and $\kappa$ is the curvature. I used the fact that $t \perp n \implies t \cdot n = 0$.
Let us assume that $\alpha$ does not pass through $x_0$ and that $\alpha$ has no inflections, i.e. $\kappa \neq 0$. Notice that if there exists a point $x_0$ and a function $\lambda$ with the property $x_0 = \alpha + \lambda n$ then $\lambda' = 0$. To prove this notice that $x_0 = \alpha + \lambda n$ gives $x_0 - \alpha = \lambda n$ and so $\lambda' = -\kappa(\lambda n) \cdot t = 0$. What this tells us is that if there is a point $x_0$ on each of the normals then it must be the same distance along each normal, i.e. be equi-distant from each point on $\alpha$. This shows that $\alpha$ follows a circular path with $x_0$ as the centre.
