In the diagram below, the circle with center $A$ is congruent to and tangent to the circle with center $B$.  A third circle is tangent to the circle with center $A$ at point $C$ and passes through point $B$.  Points $C$, $A$, and $B$ are collinear.  The line segment $\overline{CDEFG}$ intersects the circles at the indicated points.  Suppose that $DE = 6$ and $FG = 9$.  Find $AG$.
[asy]
unitsize(5);
pair A = (-9 sqrt(3), 0);
pair B = (9 sqrt(3), 0);
pair C = (-18 sqrt(3), 0);
pair D = (-4 sqrt(3) / 3, 10 sqrt(6) / 3);
pair E = (2 sqrt(3), 4 sqrt(6));
pair F = (7 sqrt(3), 5 sqrt(6));
pair G = (12 sqrt(3), 6 sqrt(6));
real r = 9sqrt(3);
draw(circle(A, r));
draw(circle(B, r));
draw(circle((B + C) / 2, 3r / 2));
draw(C -- D);
draw("$6$", E -- D);
draw(E -- F);
draw("$9$", F -- G);
dot(A);
dot(B);
label("$A$", A, plain.E);
label("$B$", B, plain.E);
label("$C$", C, W);
label("$D$", D, dir(160));
label("$E$", E, S);
label("$F$", F, SSW);
label("$G$", G, N);
[/asy]
1. **Identify the circles and their properties:**
   - Let $\omega_A$ be the circle centered at $A$.
   - Let $\omega_B$ be the circle centered at $B$.
   - Let $\omega$ be the third circle tangent to $\omega_A$ at $C$ and passing through $B$.
   - Since $\omega_A$ and $\omega_B$ are congruent and tangent, their radii are equal, say $r$.

2. **Understand the geometric configuration:**
   - Points $C$, $A$, and $B$ are collinear.
   - $\overline{BC}$ is a diameter of $\omega$, implying $BC = 2r$.
   - The line segment $\overline{CDEFG}$ intersects the circles at the indicated points.

3. **Determine the distances:**
   - Given $DE = 6$ and $FG = 9$.
   - Since $BF$ is the perpendicular bisector of $\overline{EG}$, $EF = 9$.

4. **Apply homothety:**
   - Consider the homothety centered at $C$ that maps $\omega_A$ to $\omega$.
   - The ratio of the homothety is $\frac{3}{2}$ (since $\omega$ has a radius $1.5r$).
   - This homothety maps $D$ to $F$, so $CF = \frac{3}{2} CD$.
   - Given $DE = 6$, $CD = 2 \times 15 = 30$.

5. **Calculate the radius $r$:**
   - Using the Power of a Point theorem at $C$ with respect to $\omega_B$:
     \[
     CE \cdot CG = CB^2 - r^2
     \]
     \[
     36 \cdot 54 = (3r)^2 - r^2
     \]
     \[
     1944 = 9r^2 - r^2
     \]
     \[
     1944 = 8r^2
     \]
     \[
     r^2 = \frac{1944}{8} = 243
     \]
     \[
     r = \sqrt{243} = 9\sqrt{3}
     \]

6. **Find $AG$:**
   - Using the Power of a Point theorem at $G$ with respect to $\omega_A$:
     \[
     GD \cdot GC = GA^2 - r^2
     \]
     \[
     24 \cdot 54 = GA^2 - 243
     \]
     \[
     1296 = GA^2 - 243
     \]
     \[
     GA^2 = 1296 + 243 = 1539
     \]
     \[
     GA = \sqrt{1539} = 9\sqrt{19}
     \]

The final answer is $\boxed{9\sqrt{19}}$