Assuming all you have are the positions of $A_L$, $A_R$, $A_N$, and $B_N$, that is, you don't even know where $O$ is in advance, this is a cute little geometry problem.
As $A_N$ and $B_N$ are at the same distance from $O$, the latter lies on the perpendicular bisector of the line segment $A_NB_N$. Also, it's clear from the picture that $O$ must lie on the line $A_LA_R$. This fixes the position of $O$. Then you can use Ilmari's answer to find the lengths of the arcs $d_L$ and $d_R$.
For an analytical solution, let $\vec u = \vec A_L - \vec A_R$ and $\vec v = \vec B_N - \vec A_N$. As $O$ lies on the line through $A_N$ parallel to $\vec u$, we can write $\vec O = \vec A_N + c \vec u$ for some scalar $c$. Also, as $O$ is equidistant from $A_N$ and $B_N$, we have $\lVert \vec O - \vec A_N \rVert^2 = \lVert \vec O - \vec B_N \rVert^2$. Substituting $\vec O = \vec A_N + c\vec u$ and using $\lVert \vec x \rVert^2 = \vec x \cdot \vec x$ for any $\vec x$, we get $(c \vec u) \cdot (c \vec u) = (c \vec u - \vec v)\cdot(c \vec u - \vec v),$ so $c = \frac12 \frac{\vec v \cdot \vec v}{\vec u \cdot \vec v}$ Finally, as $OA_NB_N$ is an isosceles triangle with sides $\lVert c \vec u \rVert$, $\lVert c \vec u \rVert$, and $\lVert \vec v \rVert$, the angle $\theta$ at $O$ satisfies $2 \sin \frac\theta2 = \frac{\lVert \vec v \rVert}{\lVert c \vec u \rVert},$ so $\theta = 2 \sin^{-1} \frac{\lVert \vec v \rVert}{2\lVert c \vec u \rVert} = 2 \sin^{-1} \frac{\vec u \cdot \vec v}{\lVert \vec u \rVert \lVert \vec v \rVert}$ and that should be you everything you need to compute the quantities in Ilmari's answer.