The segment can be parameterized as
$$
u(\lambda) = (1-\lambda) u_1 + \lambda u_2 \quad (\lambda \in [0, 1])
$$
where $u_i = (x_i, y_i)$ are the end points of the segment. If $\lambda \not\in I$ we are still on the line, but outside the segment.
The yellow bounding box is the union of the discs with center at $u(\lambda)$ and radius $r$:
$$
BB = \bigcup_{\lambda \in [0, 1]}
\left\{ u \mid \lVert u - u(\lambda) \rVert \le r \right\}
$$
So we know $u\in BB$ if we can find a $\lambda \in I = [0,1]$ with
$$
d(\lambda) = \lVert u - u(\lambda) \rVert \le r
$$
or
$$
0 \le q(\lambda) = \lVert u - u(\lambda) \rVert^2 \le r^2
$$
where
\begin{align}
q(\lambda) &= (u - (1-\lambda)u_1 - \lambda u_2)^2 \\
&= (u - u_1 + \lambda(u_1 - u_2))^2 \\
&= \underbrace{(u - u_1)^2}_c +
2\lambda \underbrace{(u - u_1)\cdot (u_1-u_2)}_b +
\lambda^2 \underbrace{(u_1-u_2)^2}_a \\
&= a \lambda^2 + 2 b \lambda + c \\
&= a (\lambda + b/a)^2 + c - (b^2/a)
\end{align}
is a quadratic function in $\lambda$, if $u_1 \ne u_2$, which we assume.
We note the dot in the expression for $b$ is a scalar product, and $a > 0$, $b \in \mathbb{R}$ and $c \ge 0$.
\begin{align}
d_{01} &= u - u_1 \\
d_{12} &= u_1 - u_2 \\
a &= d_{12}^2 \\
b &= d_{01} \cdot d_{12} \\
c & = d_{01}^2
\end{align}
As $a > 0$ we know that the parabola is upwards open, $q$ has a minimum. Left to the minimum it decreases, right to the minimum it increases.
From the above form (or by inspecting where the derivative vanishes) we can read that the minimum is at
$$
(\lambda^*, m) = (-b/a, c - (b^2 / a))
$$
Case 1:
If $\lambda^* \in I$, then our test if $u \in BB$ consists of
$$
m \le r^2
$$
Case 2:
If $\lambda^* < 0$, the minimum is to the left of $I$, and $q$ increases on $I$. So the minimum on $I$ is at $\lambda = 0$ and we test for
$$
q(0) = c \le r^2
$$
Case 3:
Else $\lambda^* > 1$, the minimum is to the right of $I$, and $q$ decreases on $I$. The minimum on $I$ is at $\lambda = 1$ and we test for
$$
q(1) = a + 2b + c \le r^2
$$
Example:
For $u_1 = (0,0)$, $u_2 = (10, 1)$ and $u = (3,4)$ we have
\begin{align}
a &= (u_1 - u_2)^2 = (0 - 10)^2 + (0 - 1)^2 = 101 \\
b &= (u - u_1) \cdot (u_1 - u_2) = (3, 4) \cdot (-10, -1) = -30 - 4 = -34 \\
c &= (u - u_1)^2 = (3,4)^2 = 9 + 16 = 25
\end{align}
This gives $\lambda^* = -b/a = 34/101 \in I$ and
$m = c - (b^2/a) = 25 - 34^2/101 = 13.55$
For $r=20$ the test would be
$$
m = 13.55 \le 20^2= 400
$$
which $u$ would pass.
