For $i=1,2,3$, let $C_i$ be the circle in the $xy$ plane with center $\mathbf{c}_i = (x_i,y_i)$ and radius $r_i$. Assume that these three circles all intersect one another (6 intersection points, in all). Then let $L_{ij}$ be the line through the intersection points of the circles $C_i$ and $C_j$; in other words, a common chord of both $C_i$ and $C_j$.
I think it's true that the three lines $L_{12}$, $L_{23}$ and $L_{31}$ all meet at a common point. Let's call this point $\mathbf{p}$. It's shown as a hollow (white) point in this picture:
Because of the symmetry of the situation, there must be some nice tidy symmetric coordinate-free expression that gives $\mathbf{p}$ in terms of $\mathbf{c}_1$, $\mathbf{c}_2$, $\mathbf{c}_3$. Presumably it will take the form $$ \mathbf{p} = \lambda_1\mathbf{c}_1 + \lambda_2\mathbf{c}_2 + \lambda_3\mathbf{c}_3 $$ for some scalars $\lambda_1$, $\lambda_2$, $\lambda_3$ with $\lambda_1 + \lambda_2+\lambda_3 = 1$. What is this expression? In other words, what are $\lambda_1$, $\lambda_2$, $\lambda_3$?
What have I tried, some of you will surely ask. Well, I set up a coordinate system with origin at $\mathbf{c}_1$, and with $x$-axis passing through the point $\mathbf{c}_2$. It's not too hard to figure out the coordinates of $\mathbf{p}$ in this coordinate system. But the answer is messy, assymmetrical, and coordinate-system-dependent. I'm looking for something prettier.
I'll change-up your notation a bit, centering circles at $A$, $B$, $C$ with respective radii $r$, $s$, $t$.
It is not too difficult to show that your point, call it $P$, is given by $$P = \frac{1}{4|\triangle ABC|}\left(\;(|A|^2-r^2)(B-C)^\perp+(|B|^2-s^2)(C-A)^\perp+(|C|^2-t^2)(A-B)^\perp\;\right)$$ where $(x,y)^\perp := (y,-x)$, and where $|\triangle ABC|$ is the signed area of $\triangle ABC$ (positive when $A\to B\to C\to A$ traces the triangle counter-clockwise).
Then, while it is not particularly easy to derive, it is relatively straightforward (if tedious) to verify, that $$P = \alpha A + \beta B + \gamma C$$ where $$\begin{align} \alpha &:= \frac{a}{8|\triangle ABC|^2} \left(\; a b c \cos A - r^2 a + s^2 b \cos C + t^2 c \cos B \;\right) \\[6pt] \beta &:= \frac{b}{8|\triangle ABC|^2} \left(\; a b c \cos B - s^2 b + t^2 c \cos A + r^2 a \cos C \;\right) \\[6pt] \gamma &:= \frac{c}{8|\triangle ABC|^2} \left(\; a b c \cos C - t^2 c + r^2 a \cos B + s^2 b \cos A \;\right) \end{align}$$ and $\alpha + \beta + \gamma = 1$.
Note. The $a$, $b$, $c$ and $\cos A$, $\cos B$, $\cos C$ in the formulas refer to the sides and angles of $\triangle ABC$, not the lengths and directions of coordinate vectors $A$, $B$, $C$.
Note. When $r=s=t$, the above reduces to $$ \alpha = \frac{a^2 b c \cos A}{8|\triangle ABC|^2} \qquad \beta = \frac{a b^2 c \cos B}{8|\triangle ABC|^2} \qquad \gamma = \frac{a b c^2 \cos C}{8|\triangle ABC|^2}$$