Let circumference 1 have a radius $r_1$ and its center at $(x_1,y_1)$; the converse is true for circumference 2 and $(x_2,y_2)$ and $r_2$. Then they define the following system of equations:
$$(x-x_1)^2 + (y-y_1)^2 = r_1^2,\\
(x-x_2)^2 + (y-y_2)^2 = r_2^2. \tag{1}$$
By solving this system for $x$ and $y$, we may get that they do not intersect at all (if the system is inconsistent), that they intersect at all their points (if the system is indeterminate), or that they intersect at two points (otherwise).
The solution that you reference comes from the following fact: if two circumferences intersect in two points (let's call them $P_{\alpha}$ and $P_{\beta}$), then both points lie in a straight segment (let's call it $P$) which crosses perpendicularly the straight line that connects the centers of both circumferences (let's call this segment $C$). Therefore, one can find the coordinates of $P_{\alpha}$ and $P_{\beta}$ following three steps:
- Find the midpoint of $C$.
- From said midpoint, find the point of $C$ at which it's crossed by $P$.
- The straight distance between the points of intersection $P_{\alpha}$ and $P_{\beta}$ and $C$.
Graphic visualization
The midpoint of $P$ is easily determined:
$$C_m=\left( \frac{x_1+x_2}{2}, \frac{y_1+y_2}{2} \right).\tag{2}$$
Now, in steps 2 and 3, we can make sense of the vectors that you asked about, $\vec{a}$ and $\vec{b}$. Note how these steps are defined by two vectors: one in the same direction as $C$ (step 2) and other in the direction of $P$ (step 3). This last vector is perpendicular to $C$. Actually, step 3 shows two vectors: we'll get to that later.
If those two steps are movements in relation to $C$, then it's natural to define two orthogonal vectors that act as a basis for these movements, $\vec{a}$ and $\vec{b}$:
$$\vec{a} = \left( \frac{x_2-x_1}{R}, \frac{y_2-y_1}{R} \right),
\vec{b} = \left( \frac{y_2-y_1}{R}, - \frac{x_2-x_1}{R} \right),\tag{3}$$
where $R$ is the length of $C$ and is used to normalize both vectors with respect to $C$. Now, step 2 can be reformulated to finding how much we have to move in $\vec{a}$ (this is, variable $a$ in the referenced answer). Conversely, step 3 would be finding how much we have to move in $\vec{b}$ in both directions (this is, variable $b$ in the referenced answer). Therefore, the locations of $P_{\alpha}$ and $P_{\beta}$ are defined by:
$$P_{\alpha}=C_m+a\cdot \vec{a} + b\cdot \vec{b},\\
P_{\beta}=C_m+a\cdot \vec{a} - b\cdot \vec{b},\tag{4}$$
where the two vectors of step 3 are implied by the $+$ and $-$ signs of the $b$ component. Now, if we express the equations of both circumferences in terms of $\vec{a}$ and $\vec{b}$:
$$\big( a + R / 2 \big)^2 + b^2 = r_1^2,\\
\big( a - R / 2 \big)^2 + b^2 = r_2^2.\tag{5}$$
This is basically a more handy way of writing $(1)$. Now, in the original answer, the author just solves for $a$ and $b$ and substitutes their values in $(4)$ (although this is implicitly as he never states $(4)$ explicitly).
The advantage of formulating the problem this way is that it allows to simplify the second grade terms that appear when you solve $(1)$. Also, it allows to better emphasize the symmetric relation between both points of intersection by means of the $+$ and $-$ signs. You can also see directly that, in the extreme case where the two circumferences intersect at a single point, then $b=0$; this is, this single point must lie on C.
