Imagine a typical drunkards walk (2D) made of steps $\ell$ each of length $L$ in any direction.
I was told that the probability distribution of each step can be written as a Dirac delta like this $$\chi(\ell) = \frac{\delta(|\ell|-L)}{2 \pi L},$$ evenly spaced around the circle.
When plotting it in 3D i get this]
https://i.stack.imgur.com/EaSS8.png
Why would the probability for each step look like this?
This isn't a proof but merely motivation/intuition for that result.
We want the drunkard to take a step that lands him at a point on a circle of radius $L$ that is as likely as any other point of that circle.
In general, what is the uniform probability distribution $f(t)$ over the interval $[a, b]$ where $b>a$? It's $f(t) = \frac{1}{b-a}$ since $\int_a^b f(t)\,dt = 1$.
Now imagine you have a circle of radius $r$. Its circumference is $2\pi r$. Further, imagine laying that circle out as an interval $[a,b]$ in one dimension, with $a=0$ and $b=2\pi r$. Then you can see that a uniform distribution over the points on that circumference is $f(r) = \frac{1}{2\pi r}$.
We want the steps to be of length $L$, however, not any arbitrary $r$. We can enforce that by choosing $f(r) = \frac{\delta(r-L)}{2\pi r}$ instead. But that is the same distribution as $f(r) = \frac{\delta(r-L)}{2\pi L}$.
A little more formally, if $\mathbf{u}$ is the vector describing a step and if $f(\mathbf{u})$ is the probability distribution of that step, then $$ \int_{\mathbb{R}^2}\,f(\mathbf{u})\,d\mathbf{u} = 1 $$
Let $\mathbf{u}$ have magnitude $r$ and angle $\theta$. Then $d\mathbf{u} = r\,dr\,d\theta$
[Why? Because we're looking for the probability distribution of steps that will make him land on an infinitesimal patch of area $r\,dr\,d\theta$ centered at the point with polar coordinates $(r,\theta)$]
and $$ \int_{\mathbb{R}^2}\,f(\mathbf{u})\,d\mathbf{u} = \int_{r\,=\,0}^{\infty}\int_{\theta\,=\,0}^{2\pi}\,f(r,\theta)\,r\,dr\,d\theta = 1 $$
The solution that is uniform in $\theta$ but restricted to $r=L\,$ is $\,f(r,\theta) = \frac{\delta{(r-L)}}{2\pi L}$.
As for why the plot looks like what you have, I think you're plotting it incorrectly. Try plotting DiscreteDelta[Abs[(x^2 + y^2)^(1/2) - L] / (2 Pi L) instead, for a fixed choice of $L$ and variable $(x,y)$.