<s>[INST] 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?
 [/INST] 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, $|\triangle T| = 0$.
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