Don't know that it's much simpler than calculating the pairwise intersections, then the distances to the third center, but the following gives a symmetric condition using complex numbers.
Let $\,a,b,c\,$ be the complex numbers associated with points $A,B,C$ in a complex plane centered at the centroid of $ABC\,$, so that $a+b+c=0\,$.
The point of intersection $z$ of the three circles (if it exists) must satisfy the $3$ equations similar to:
$$
|z-a|^2=R_A^2 \;\;\iff\;\;(z-a)(\bar z - \bar a) = R_A^2 \;\;\iff\;\;|z|^2 - z \bar a - \bar z a + |a|^2 = R_A^2 \tag{1}
$$
Writing $(1)$ for $a,b,c$ and summing the $3$ equations up:
$$
\require{cancel}
3\,|z|^2 - \cancel{z \sum_{cyc} \bar a} - \bcancel{\bar z \sum_{cyc} a} + \sum_{cyc}|a|^2 = \sum_{cyc} R_A^2 \;\;\implies\;\; |z|^2 = \frac{1}{3}\left(\sum_{cyc} R_A^2-\sum_{cyc}|a|^2\right) =R^2 \tag{2}
$$
Substituting $(2)$ back into each of $(1)\,$:
$$
-|z|^2 + z \bar a + \bar z a - |a|^2 = - R_A^2 \;\;\iff\;\; z \cdot \bar a + \bar z \cdot a = |a|^2+R^2-R_A^2 \tag{3}
$$
Considering $(3)$ as a system of linear equations in $z, \bar z\,$, the condition for it to have solutions is:
$$
\left|
\begin{matrix}
\;\bar a \;&\; a \;&\; |a|^2+R^2-R_A^2\; \\
\;\bar b \;&\; b \;&\; |b|^2+R^2-R_B^2\; \\
\;\bar c \;&\; c \;&\; |c|^2+R^2-R_C^2\;
\end{matrix}
\right| \;\;=\;\; 0
$$
Given disjoint circles, and unequal radii, the locus of centers comprises two hyperbolas. Begin by intersecting the axis with the both circles. Let it intersect circle $A$ at $A_1$ and $A_2$, and circle $B$ at $B_1$ and $B_2$, as shown here, where $A_1$ and $B_1$ are between the two centers.
Let $K$ be the midpoint of $A_2B_2$, and $L$ the midpoint of $A_1B_1$. Let $P$ be the center of a circle externally tangent to both or internally tangent to both. This relation follows:
$(PA - PB)^2 = (r_a-r_b)^2$
The locus of $P$ is a hyperbola with foci $A$ and $B$. Points $K$ and $L$ both satisfy the condition for $P$, and they lie on the axis, so those are the vertices.
Now start again. Let $M$ be the midpoint of $A_2B_1$, and $N$ the midpoint of $A_1B_2$. Let $Q$ be the center of a circle externally tangent to one of the given circles and internally tangent to the other. This relation follows:
$(QA - QB)^2 = (r_a+r_b)^2$
The locus of $Q$ also is a hyperbola with foci $A$ and $B$. This time the vertices are at $M$ and $N$.
Other cases to investigate would be intersecting circles or congruent circles.
Best Answer
You can find a formula for the area of intersection between two circles here. Using only this formula, you can find the area you are looking for.
In the figure above I've drawn the scenario you describe, with names for the various areas which make up the intersection area between the circles. If we denote the Area of Intersection between the two circles with radii $r_1$ and $r_2$ as $AI(r_1,r_2)$, we see that: $$A_2 + 2 \cdot A_1 = AI(r_c,r_a) - AI(r_d, r_a) = C_1$$ $$A_3 + 2 \cdot A_1 = AI(r_a,r_c) - AI(r_b, r_c) = C_2$$ $$A_2 + A_3 + 2 \cdot A_1 = AI(r_c,r_a) - AI(r_d, r_b) = C_3$$ Therefore, $$C_1 + C_2 - C_3 = 2 \cdot A_1$$ or $$A_1 = \frac{C_1 + C_2 - C_3}{2}$$