Three Circles Meeting at One Point We have three triples of points on the plane, that is, $X=\{x_1, x_2, x_3\}$, $Y=\{y_1, y_2, y_3 \}$, and $Z=\{z_1, z_2, z_3\}$, where $x_i, y_i, z_i$ are points on the plane. I was wondering if there is a simple (or not-so-simple) algebraic relation satisfied by the coordinates of the 9 points if we know that three circles determined by $X, Y, Z$ pass through a single point. 

Let $\triangle ABC$ have edge-lengths $a=|BC|$, $b=|CA|$, $c=|AB|$. Suppose circles of radius $u$, $v$, $w$ about respective points $A$, $B$, $C$ meet at a common point $P$.
Coordinatizing, we can write $A=(0,0)$, $B=(c,0)$, $C=(b\cos A, b\sin A)$, $P=(p,q)$, so that
$$\begin{align}
u^2 &= p^2 + q^2 &(1)\\
v^2 &= (p-c)^2 + q^2 = p^2 + q^2 - 2 p c + c^2 &(2)\\
w^2 &= (p-b \cos A)^2 + (q-b\sin A)^2= p^2 + q^2-2b p \cos A - 2 b q \sin A + b^2 &(3)
\end{align}$$
Use $(1)$ to eliminate $p^2+q^2$ from $(2)$ and $(3)$:
$$\begin{align}
u^2 - v^2 + c^2 &= 2 p c &(4)\\
u^2 - w^2 + b^2 &= 2 b p \cos A + 2 b q \sin A &(5)
\end{align}$$
Then, solve the system $(4)$, $(5)$ for $p$ and $q$. Substituting this solution into $(1)$, replacing $\sin^2 A$ with $1-\cos^2 A$, and then writing $\cos A = \frac{1}{2bc}(-a^2+b^2+c^2)$, gives this polynomial condition for the concurrence of the three circles:
$$\begin{align}
 0 &= a^2 b^2 c^2 + a^2 u^4 + b^2 v^4 + c^2 w^4 \\
&\quad- \left(\;a^2 u^2 + v^2 w^2\;\right)\left(-a^2 + b^2 + c^2\;\right) \\
&\quad- \left(\;b^2 v^2 + w^2 u^2\;\right)\left(\phantom{-}a^2 - b^2 + c^2\;\right) \\
&\quad- \left(\;c^2 w^2 + u^2 v^2\;\right)\left(\phantom{-}a^2 + b^2 - c^2\;\right) 
&(\star)\end{align}$$

To answer your question, you "only" need to determine centers $A$, $B$, $C$, distances, $a$, $b$, $c$, and radii $u$, $v$, $w$ from your point-sets $X$, $Y$, $Z$, and then make appropriate substitutions into $(\star)$. Doing this symbolically is straightforward, if tedious ---see, for instance, the MathWorld "Circumcircle" article--- and the result will be an enormous polynomial. As a practical matter, it's probably best to compute the various elements numerically, and then substitute. 
