Curves And Surfaces Exercise (6) of Chapter 1

Question

I am currently reading the book "CURVES AND SURFACES" [ SECOND EDITION 2009 ] by Ros and Montiel.

I am having trouble to solve and understand Exercise (6) of Chapter 1.
It says the following Exercise Statement :

Show that a curve $ \alpha: I \rightarrow \mathbb{R}^2 $ is a segment of a straight line or an arc of a circle if and only if all its tangent lines are equidistant from a given point in the plane.

Also the Hint given says :

Assume that $a\in \mathbb{R}^2 $ is the point from which all the tangent lines of the curve $ \alpha: I \rightarrow \mathbb{R}^2 $ are equidistant. Let $ s \in I $ be an arbitrary parameter value. The tangent line to the curve at $ s $ passes through $ \alpha(s) $ and is perpendicular to $ N(s)$ , the normal vector at $ s$.
We have
$$ \langle \alpha(s) - a, N(s) \rangle = c, $$
where $ \langle \cdot, \cdot \rangle$ denotes the inner product, $ c \in \mathbb{R} $ is a constant, and $ s \in I $.
Taking derivatives and using the Frenet equations, we get that for each $ s \in I$ , either $ k(s) = 0 $ or $ 1 + ck(s) = 0 $.
Since $ k $ is continuous, it follows that $ k $ is either identically zero or identically $ -1/c $ over $ I $.

My Problem : I do not understand how to arrive at the equation
$$ \langle\alpha(s) - a, N(s) \rangle = c. $$

I think it makes the additional assumption that the distance between the tangent line and the point $a$ is realized at $\alpha(s)$, which I do not see why it should be the case.
It could be realized at a point $\alpha(s) + \lambda(s) \cdot \alpha'(s)$ for some $\lambda(s) \in \mathbb{R}$ ...

Answer 1

Denote by $\beta(s)$ the point on the tangent where the minimum distance to $a$ is achieved. We have

$$c=\langle \beta(s) - a, N(s) \rangle = \langle \beta(s) -\alpha(s) + \alpha(s)- a, N(s) \rangle = \langle \alpha(s) - a, N(s) \rangle$$ as $\langle \beta(s) - \alpha(s), N(s) \rangle = 0$. Proving the desired result.

Answer 2

No, you don't need that additional assumption. As you observed, without it we obtain an equation of the form $$\langle\alpha(s)+\lambda(s)\alpha'(s)-a, N(s)\rangle= \pm c$$ for some constant $c \geq 0$. But now just note that $$\langle \alpha'(s), N(s)\rangle =0$$ which yields exactly $$\langle\alpha(s)-a, N(s)\rangle= \pm c$$ where the sign $\pm$ is uniquely determined by continuity.