<s>[INST] The lines that connect the intersections of the three circles around a triangle meet at the same point 
For any $\Delta ABC$, construct three circles with radius $AB,CB,AC$ centered at $A,B,C$ respectively. Then, connect the intersections of the circles as shown in the picture. The lines are concurrent and meet at point $O$. I am curious which center this is among the thousands of the known triangle centers. 
I know that I can find it by calculating the barycentrics and the trillinears, but could you teach me how?
 [/INST] The lines through two points of intersection of two circles is called the radical axis; the axis is actually defined for any two non-concentric circles, and the radical axis theorem states that the axes for each pair of three given circles are concurrent. So, that property of the configuration in question is unsurprising.
As @darij notes in a comment, the point described is not a triangle center, as its description is not properly symmetric in the elements of the triangle. Nevertheless, I'll walk through finding the barycentric and trilinear coordinates, since OP wants to see how that's done.

Defining $a := |BC|$, $b := |CA|$, $c := |AB|$ as is typical, we can assign these (Cartesian) coordinates
$$A = (0,0) \qquad B=(c,0) \qquad C = (b \cos A, b \sin A)$$
Then we have
$$\begin{align}
\bigcirc A:\quad x^2 + y^2 &= c^2 \\
\bigcirc B:\quad x^2 + y^2 &= b^2 - 2 b c \cos A + 2 xc \quad\to\quad x^2+y^2= a^2 - c^2+ 2 x c \\
\bigcirc C:\quad x^2 + y^2 &= 2 b ( x\cos A + y\sin A) 
\end{align}$$
The equations of the radical axes of two circles (intersecting or not) are dead-simple to calculate: simply combine the circle equations to eliminate $x^2$ and $y^2$ terms; here, simple subtraction works. Solving any two of the resulting equations, and manipulating appropriately, gives
$$P = 
\left(\frac{-a^2 + 2 c^2}{2 c}, \frac{-a^4 + a^2 b^2 + 3 a^2 c^2 - 2 b^2 c^2}{
 4 b c^2 \sin A}\right)$$
Writing $$P = \frac{\alpha A + \beta B + \gamma C}{\alpha+\beta+\gamma}$$ we can solve for $\alpha$, $\beta$, $\gamma$ to get
$$\begin{align}
\alpha:\beta:\gamma \quad=\quad &b^4 - 3 a^2 b^2 - \phantom{3}b^2 c^2 + 2 c^2 a^2  \\[4pt]
:\; &c^4 + 2 a^2 b^2 - 3 b^2 c^2 - \phantom{3}c^2 a^2  \\[4pt]
:\; &a^4 - \phantom{3}a^2 b^2 + 2 b^2 c^2 - 3 a^2 c^2 
\end{align}$$
These are the barycentric coordinates of $P$. The trilinear coordinates of $P$ are $\alpha/a : \beta/b :\gamma/c$.
Given barycentrics $\alpha:\beta:\gamma$, we could search the Encyclopedia of Triangle Centers using a "normalized" value generated from specific edge-lengths $a=6$, $b=9$, $c=13$. In this case, we have
$$\frac{2|\triangle ABC|}{\alpha+\beta+\gamma}\;\frac{\alpha}{a} = 3.26441\ldots$$
This number doesn't not appear in the table, but we already knew not to expect it, since $P$ is not a symmetrically-defined triangle center. $\square$

For an example of a related point that is a Triangle Center, take $\bigcirc A$, $\bigcirc B$, $\bigcirc C$ to have respective radii $a$, $b$, $c$. (See the symmetry in the definition?) Using the techniques above, we find that the point of concurrency of the three circles is 
$$P = \left( c - b \cos A, \frac{b - 2 c \cos A + b \cos A^2}{\sin A}\right)$$
so that it has barycentric coordinates
$$\begin{align}
\alpha:\beta:\gamma \quad=\quad &-3 a^4 + \phantom{3} b^4 + \phantom{3}c^4 + 2 a^2 b^2 + 2 a^2 c^2 - 2 b^2 c^2 \\[4pt]
:\;&\phantom{\,-3}a^4 - 3 b^4 + \phantom{3}c^4 + 2 a^2 b^2 - 2 a^2 c^2 + 2 b^2 c^2 \\[4pt]
:\;&\phantom{\,-3}a^4 + \phantom{3}b^4 - 3 c^4 - 2 a^2 b^2 + 2 a^2 c^2 + 2 b^2 c^2
\end{align}$$
The corresponding lookup number at the ETC is $19.2413\ldots$, which we find leads to the De Longchamps Point, $X(20)$. $\square$
</s>