THIS ANSWER GIVES YOU THE FAMILY OF CIRCLES THROUGH TWO GIVEN POINTS.
The way I would work is something like this... if the two points $P_1(x_1,y_1)$ and $P_2(x_2,y_2)$ are on your circle then the line segment $[P_1P_2]$ is a chord. Suppose that the distance $|P_1P_2|=2d$.
Now the perpendicular from the centre of a circle to a chord bisects the chord. We can get the equation of this perpendicular bisector because we have that the midpoint of $[P_1P_2]$ is on it... $((x_1+x_2)/2,(y_1+y_2)/2)$. The slope of the perpendicular bisector, $m$, satisfies $$m\cdot m_{[P_1P_2]}=-1,$$
so is
$$m=-\frac{x_2-x_1}{y_2-y_1},$$
so the perpendicular bisector has equation
$$y=-\frac{x_2-x_1}{y_2-y_1}x+\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2},$$
briefly $k=mh+c$ where $k\sim y$, $h\sim x$ and
$$m=-\frac{x_2-x_1}{y_2-y_1},$$
and
$$c=\frac{x_2-x_1}{y_2-y_1}\frac{x_1+x_2}{2}+\frac{y_1+y_2}{2}$$
Here $h$ is free --- each $h$ gives you a different centre $(h,k)=(h,mh+c)$.
By drawing a picture you will see a RAT triangle with vertices at, say the midpoint of $[P_1P_2]$, the centre $(h,mh+c)$ and $P_1$. Using Pythagoras we have that
$$r^2=d^2+\left(\frac{x_1+x_2}{2}-h\right)^2+\left(\frac{y_1+y_2}{2}-(mh+c)\right)^2$$
So we have that the circles in question are:
$$\{(x-h)^2+(x-(mh+c))^2=r^2:h\in\mathbb{R}\},$$
where each of $m$ and $c$ are functions of $P_1,\,P_2$ and --- once $h$ is chosen --- $r$ is a function of $P_1$, $P_2$ and $h$.
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:
$$
\left|
\begin{matrix}
\;\bar a \;&\; a \;&\; |a|^2+R^2-R_A^2\; \\
\;\bar b \;&\; b \;&\; |b|^2+R^2-R_B^2\; \\
\;\bar c \;&\; c \;&\; |c|^2+R^2-R_C^2\;
\end{matrix}
\right| \;\;=\;\; 0
$$
Best Answer
If the two circles meet at one point the distance between their centres is equal to either the sum or the difference of the radii. If the distance between the centres is smaller than the difference of the radii or larger than the sum of the radii, they do not meet. The circles meet at two points otherwise.