The radius of a set $S \in \mathbb{R}^L$ is defined as: $$\operatorname{rad}(S) = \inf_{x \in \mathbb{R}^L} \sup_{y \in S} \|x-y\|$$
I interpret the definition as follows: the radius of $S$ is related to the smallest element in $\mathbb{R}^L$ (not necessarily in $S$) that leads to the highest (Euclidean) distance to an element $y \in S$.
Could someone please provide a (geometrical) intuition for this definition?
Geometrical intuition is: pick an element $x\in\mathbb R^L$ and let $r_x$ be the radius of the smallest (closed) circle centered in $x$ that contains the whole $S$. One can see is that $||x-y||\le r_x$ for all $y\in S$, so $r_x$ is an upper bound of all the values $||x-y||$. As we wanted it to be the smallest such bound, we have:
$$r_x=\sup_{y\in S}||x-y||$$
So far so good - for every $x$ we've found the smallest (closed) circle centered in $x$ containing $S$. Now, let's move $x$ across $\mathbb R^L$, to find a point $x$ at which this radius is the smallest. And, in fact, we are not looking for the exact $x$ here - we are just looking for that smallest radius. Thus, we take:
$$r=\inf_{x\in\mathbb R^L}r_x=\inf_{x\in\mathbb R^L}\sup_{y\in S}||x-y||$$
and call that number the "radius" of $S$.
If, for example, $S$ is itself a circle with radius $\rho$, and $x$ is at distance $d_x$ from the centre, to cover the whole $S$ with the circle centered in $x$, you need the radius of the covering circle to be $r_x=d_x+\rho$ to reach the furthermost point of $S$. Obviously, across all points $x$ this will be the smallest if $x$ is the centre of $S$ and $d_x=0$, which means that the radius of $S$ will be $0+\rho = \rho$. This means that, for circles, the radius in this new sense corresponds to the radius in the usual sense.
You can check (as an exercise) that: