Determine if 3 circles intersect at a common point Given three circles centered at points $A$, $B$, and $C$ with non-zero radii of lengths $R_A$, $R_B$ and $R_C$. Where the centers of the circles form a valid triangle, and where the distance between any two centers is less than or equal to the sum of their corresponding radii.
Question: Is it possible to determine if all three circles intersect at a common point using a calculation simpler than first determining the pair of intersection points between two pairs of circles then determining if any of the intersection points are equal?
 A: 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:
$$
\left| 
\begin{matrix}
\;\bar a \;&\; a \;&\; |a|^2+R^2-R_A^2\; \\
\;\bar b \;&\; b \;&\; |b|^2+R^2-R_B^2\; \\
\;\bar c \;&\; c \;&\; |c|^2+R^2-R_C^2\;
\end{matrix}
\right| \;\;=\;\; 0
$$
A: You can calculate the pair of intersection between, say, circles $A$ and $B$, then calculate the distance of each such intersection to circle $C$. There is a triple intersection if and only if one of the distances is $R_C$.
EDIT: Well, consider $\triangle ABC$ and suppose there is a triple intersection, that is, some point $P$ with $PA=R_A$, $PB=R_B$ and $PC=R_C$. Let $E_A$ be the side of $\triangle ABC$ opposite vertex $A$ and similarly for $B$ and $C$.
Suppose without loss of generality that $R_A\leq R_B\leq R_C$. The triangle inequality implies that the following must hold:
\begin{align}
R_C-R_B\leq E_A\leq R_C+R_B\\
R_C-R_A\leq E_B\leq R_C+R_A\\
R_B-R_A\leq E_C\leq R_B+R_A
\end{align}
By hypothesis, the right hand inequalities do hold, but if any of the left hand ones do not, you can already rule out the possibility of a triple intersection.
I might yet improve on this later.
A: If we coordinatize, say, with
$$A = (0,0) \qquad B = (c, 0) \qquad c = (b\cos A,b\sin A)$$
and take suppose circles $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ (of respective radii $r_A$, $r_B$, $r_C$) meet at a point $P = (x,y)$, then we have three equations in two unknowns $x$ and $y$:
$$\begin{align}
x^2 + y^2 &= r_A^2 \\
x^2 + y^2 &= r_B^2 + 2 c x - c^2 \\
x^2 + y^2 &= r_C^2 + 2 b x \cos A x + 2 b y \sin A - b^2 
\end{align}$$
We can eliminate $x$ and $y$ from these equations, leaving this relation:
$$\begin{align} a^2 b^2 c^2 + a^2 r_A^4 + b^2 r_B^4 + c^2 r_C^4 &= \left( a^2 r_A^2 + r_B^2 r_C^2 \right) \left(-a^2 + b^2 + c^2 \right) \\
&+ \left( b^2 r_B^2 + r_C^2 r_A^2 \right) \left(\phantom{-}a^2 - b^2 + c^2 \right)\\
&+ \left( c^2 r_C^2 + r_A^2 r_B^2 \right) \left(\phantom{-}a^2 + b^2 - c^2 \right)
\end{align} \tag{1}$$
The right-hand side seems to want to be re-written with cosines ...
$$\begin{align} a^2 b^2 c^2 + a^2 r_A^4 + b^2 r_B^4 + c^2 r_C^4 &= 2 b c \cos A \left( a^2 r_A^2 + r_B^2 r_C^2 \right) \\
&+ 2 c a \cos B \left( b^2 r_B^2 + r_C^2 r_A^2 \right)\\
&+ 2 a b \cos C \left( c^2 r_C^2 + r_A^2 r_B^2 \right) 
\end{align} \tag{2}$$ 
... but this doesn't seem a great deal better. Perhaps if we use the Law of Sines to write
$$a = 2 r \sin A \qquad b = 2 r \sin B \qquad c = 2 r \sin C$$
where $r$ is the circumradius of $\triangle ABC$. With a little effort, we find this form for the relation:
$$\begin{align}
\frac{1}{64} \left(\;\begin{array}{c} r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C \\
- 8 r^2 \sin A \sin B \sin C\end{array} \;\right)^2
= \frac{1}{16}&(\phantom{-}r_A \sin A + r_B \sin B + r_C \sin C ) \\
\cdot &(-r_A \sin A + r_B \sin B + r_C \sin C )\\
\cdot &(\phantom{-}r_A \sin A - r_B \sin B + r_C \sin C )\\
\cdot &(\phantom{-}r_A \sin A + r_B \sin B - r_C \sin C )
\end{align} \tag{3}
$$
The curious fractional coefficients are there to help us recognize the right-hand side as Heron's Formula for the square of the area of a triangle with side-lengths $r_A \sin A$, $r_B \sin B$, $r_C \sin C$. More precisely, when (and only when) the right-hand side of $(3)$ is non-negative, it gives the square of the area of the triangle with those side-lengths; when (and only when) the right-hand side is negative, those side-lengths fail to form a valid triangle. (Note: I consider a degenerate triangle of area $0$ to be valid.) 
Since the left-hand side of $(3)$ is necessarily non-negative, we deduce that 

$\bigcirc A$, $\bigcirc B$, $\bigcirc C$, with radii $r_A$, $r_B$, $r_C$, concur at a point only if $r_A \sin A$, $r_B \sin B$, $r_C \sin C$ are the edges of a valid triangle (ie, they satisfy the Triangle Inequality).

That gives you a way to weed-out bad candidates. To know for sure that the three circles concur, you'd need to check the full equality of $(3)$. (Is that "simpler" than the strategy you mentioned? I'm not sure.)

Note that $8 r^2 \sin A\sin B\sin C = 2 a b \sin C = 4 |\triangle ABC|$. If we call the "valid triangle" referenced above, say, $\triangle T$, then we can write $(3)$ as

$$r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C \pm 8 |\triangle T|\;=\; 4 |\triangle ABC|
\tag{4}$$

To get at the behavior of the "$\pm$", consider $P$ at distance $p$ from the circumcenter of $\triangle ABC$ (and at distances $r_A$, $r_B$, $r_C$ from $A$, $B$, $C$, respectively). With the help of Mathematica, we get
$$\begin{align}
r_A^2 \sin 2A + r_B^2 \sin 2B + r_C^2 \sin 2C &= 4 (r^2 + p^2) \sin A \sin B \sin C \\
8|\triangle T| &= 4 |r^2-p^2|\sin A \sin B \sin C
\end{align}$$
We see, then, that "$\pm$" must be "$+$" when $r > p$ (that is, when $P$ is inside the circumcircle) and "$-$" when $r < p$ (when $P$ is outside the circumcircle); for $P$ on the circumcircle, $|\triangle T| = 0$.
