Well, the problem is a question in Montiel's book. How to prove that a planar curve $\alpha$ such that all osculating circles intersects a given point is actually a circle (or a part of it)?
I've tried to use the expression $\alpha(t)+\frac{1}{k(s)}(N(s)+v(s))$ where $v(s)$ is a unit vector.
Using that $\alpha(s)=aT(s)+bN(s)$, I could only proof that (if $k'\neq 0 $) $$k=\frac{-b}{a^2+b^2}$$ and $$v'=-k'\alpha.$$
The center of the osculating circle is given by $C=P+\rho\mathbf{N}$, where all quantities depend on $s$ and $\rho=1/\kappa$ is the radius of curvature.
If the fixed point $Q$ belong to all osculating circles, then
$$
(Q-C)^2=\rho^2
$$
and differentiating (taking into account $Q'=\mathbf{0}$)
$$
(Q-C)\cdot(-C')=\rho\rho'\tag1.
$$
We can obtain $C'$ from the previous expression for $C$:
\begin{align}
C'&=P'+\rho\mathbf{N}'+\rho'\mathbf{N}\\
&=\mathbf{T}+\rho(-\kappa\mathbf{T})+\rho'\mathbf{N}=\rho'\mathbf{N}
\end{align}
where we have used the Frenet formula $\mathbf{N}'=-\kappa\mathbf{T}+\tau\mathbf{B}$ with $\tau=0$ (because the curve is a plane curve).
The eq. $(1)$ becomes
$$
-(Q-C)\cdot\mathbf{N}\rho'=\rho\rho'
$$
The relation is true if $\rho'=0$ and then also $C'=0$, so that the osculating circle is constant (does not depend on $s$), and the curve
$$
P=C-\rho\mathbf{N}
$$
is a circle or an arc of it.
If by contradiction $\rho'\neq0$ then
$$
\rho=-(Q-C)\cdot\mathbf{N}=(C-Q)\cdot\mathbf{N}=
\rho\cos\theta\quad\implies\quad\theta=0
$$
where $\theta$ is the angle between $\mathbf{N}$ and the radius $C-Q$, i.e.
$$
C-Q=\rho\mathbf{N}=C-P\quad\implies\quad P=Q=\text{const}
$$
