The objects of your universe are closed segments of positive length, closed half lines, and full lines in the plane, in short: closed one-dimensional convex sets in ${\mathbb R}^2$. Denote the set of these objects by ${\cal O}$. The intersection of two such objects is one of the following: empty, a single point, or again an element of ${\cal O}$.
Any $S\in{\cal O}$ can be extended to a full line $\ell_S$, the carrier line of $S$, and has a direction which can be represented by a unit vector ${\bf u}_S$, whereby ${\bf u}_S$ is determined only up to a factor $\pm1$.
Given two objects $S_1$, $S_2\in{\cal O}$, let $\ell_i$ be their carrier lines and ${\bf u}_i$ their directions. At the top level we then have the following three cases:
(a) $\quad {\bf u}_1\ne\pm {\bf u}_2\>$,
(b) $\quad {\bf u}_1=\pm {\bf u}_2\>,\ $ but $\ \ell_1\ne\ell_2$
(c) $\quad \ell_1=\ell_2\>$.
In case $(a)$ the two carrier lines intersect in a single point ${\bf p}$, and we have to test whether ${\bf p}\in S_1\cap S_2$. The exact form of this test depends on the way the objects are presented.
In case (b) the two carrier lines are parallel and disjoint, hence $S_1\cap S_2$ is empty.
In case (c) the common carrier line $\ell$ has a representation of the form
$$\ell:\quad t\mapsto{\bf p}+t{\bf u}\qquad(-\infty<t<\infty)\ .$$
The sets $J_i:=\ell^{-1}(S_i)$ are then closed intervals of ${\mathbb R}$, and it remains to find the intersection of these intervals.
There is absolutely nothing fuzzy, or even illogical, in this story.
You should consider this problem as a problem with subsets of $[0,2\pi)$ (or equivalently, of $[0,360)$ using degrees). Let arc $A=[a,b)$ and $B=[c,d)$ such that $a<c<b<d$. Then $A=[a,c)\cup[c,b)$ and $B=[c,b)\cup[b,d)$ and if we 'add' (take the union) them we get $A\cup B=[a,c)\cup[c,b)\cup [c,b)\cup[b,d)=[a,d)$ which is what we wanted.
Best Answer
One way to approach is this problem is to think circles rather than arcs:
First check if the two large circles intersect at all. This can be done by comparing the distance between the centers ($d$) with the sum of radii ($r_1 + r_2$). Let's suppose $r_1 \ge r_2$
If $d > r_1 + r_2$ then the circles are outside each other and don't have any intersection points, and therefore neither do the arcs.
If $d = r_1 + r_2$ then the circles are outside each other and touch at one point only, which is on the line connecting their centers.
If $r_1 - r_2 < d < r_1 + r_2$ then the circles meet at two points.
If $d = r_1 - r_2$ then the smaller circle is in the larger circle and touches it at only one point (analysis of the case where the radii are equal and the circles completely overlap is left as an exercise to the interested reader)
If $d < r_1 - r_2$ then the smaller circle is in the larger circle and the two circles don't have any intersection points, therefore neither do the arcs.
Next, find the meeting points (if any) of the circles. This can be done by solving the system of equations of the two circles. This step gives up to 2 points.
Next, for each intersection point identified in the previous step, determine if it lies on the given arcs. This can be done through the following steps for each arc:
If the point of intersection of the circles is on both arcs then you have an answer.