We are considering a $C^2$ curve $c: \mathbb{R}\to \mathbb{R}^2$ that is parametrized by arc-length (so unit speed) and satisfies det$\begin{pmatrix}c_1& c_2&\\ c_1''& c_2'' \end{pmatrix} = 0$. I would like to show that $c$ is either a line or a circle centered at the origin.
If we can reduce to the cases that $c''$ is either identically zero or non-zero everywhere, then we are done. Is it possible to reduce to these cases from the given assumptions? Otherwise we just have that $c$ is constructed piece-wise by lines and circular arcs.
From the determinant condition we have the proportion $c_1:c_1''=c_2:c_2''$. If $c_1=0$ or $c_2=0$ we get the coordinate axes, which are lines. If neither is $0$ and $c_1''=0$ then $c_2''=0$ for the proportion to hold, and integrating twice we get general lines.
Now consider the case when nothing in the proportion is $0$. From the natural parametrization condition we have that $c_1'^2+c_2'^2=1$, and differentiating $c_1'c_1''+c_2'c_2'' = 0$. The proportion then implies $c_1'c_1+c_2'c_2 = 0$ (replacing values by proportional ones), and integrating $c_1^2+c_2^2=\text{const}$. But this is the equation of a circle centered at the origin.
The converse, that naturally parameterized lines and circles satisfy the determinant condition, is easy to check by direct differentiation.