To solve the problem, we need to find the value of \( \cos 2\theta \) where \( \theta \) is the angle \( ACB \) in triangle \( ABC \) with vertices \( A(0, 3) \), \( B(0, 12) \), and \( C(x, 0) \). The circumcircle of triangle \( ABC \) touches the x-axis.
### Step-by-Step Solution:
1. **Identify Points**:
- Let \( A = (0, 3) \), \( B = (0, 12) \), and \( C = (x, 0) \).
2. **Circumcircle Touching the x-axis**:
- Since the circumcircle touches the x-axis, the center of the circumcircle must lie directly above the x-axis at a distance equal to the radius. The center will be at \( (h, k) \) where \( k \) is the radius.
3. **Finding the Center**:
- The center of the circumcircle can be found using the formula for the circumradius \( R \) of triangle \( ABC \):
\[
R = \frac{abc}{4K}
\]
where \( a, b, c \) are the lengths of the sides opposite to vertices \( A, B, C \) respectively, and \( K \) is the area of triangle \( ABC \).
4. **Calculate Lengths of Sides**:
- Length \( AB = 12 - 3 = 9 \)
- Length \( AC = \sqrt{(0 - x)^2 + (3 - 0)^2} = \sqrt{x^2 + 9} \)
- Length \( BC = \sqrt{(0 - x)^2 + (12 - 0)^2} = \sqrt{x^2 + 144} \)
5. **Calculate Area \( K \)**:
- The area \( K \) of triangle \( ABC \) can be calculated as:
\[
K = \frac{1}{2} \times \text{base} \times \text{height} = \frac{1}{2} \times 9 \times x = \frac{9x}{2}
\]
6. **Substituting into Circumradius Formula**:
- Substitute \( a = 9 \), \( b = \sqrt{x^2 + 9} \), \( c = \sqrt{x^2 + 144} \), and \( K = \frac{9x}{2} \) into the circumradius formula:
\[
R = \frac{9 \cdot \sqrt{x^2 + 9} \cdot \sqrt{x^2 + 144}}{4 \cdot \frac{9x}{2}} = \frac{\sqrt{x^2 + 9} \cdot \sqrt{x^2 + 144}}{2x}
\]
7. **Setting the Center**:
- The center of the circumcircle must be at \( (h, R) \) where \( R \) is the distance from the x-axis. Since the circumcircle touches the x-axis, the y-coordinate of the center equals the radius \( R \).
8. **Finding \( \cos 2\theta \)**:
- We can use the relationship:
\[
\cos 2\theta = 2\cos^2\theta - 1
\]
- To find \( \cos \theta \), we can use the triangle's properties:
\[
\tan \theta = \frac{3}{x} \quad \text{(from point A to C)}
\]
- Therefore,
\[
\cos \theta = \frac{x}{\sqrt{x^2 + 9}}
\]
9. **Final Calculation**:
- Substitute \( \cos \theta \) into the formula for \( \cos 2\theta \):
\[
\cos 2\theta = 2\left(\frac{x}{\sqrt{x^2 + 9}}\right)^2 - 1 = 2\frac{x^2}{x^2 + 9} - 1
\]
- Simplifying gives:
\[
\cos 2\theta = \frac{2x^2 - (x^2 + 9)}{x^2 + 9} = \frac{x^2 - 9}{x^2 + 9}
\]
### Conclusion:
The value of \( \cos 2\theta \) is \( \frac{x^2 - 9}{x^2 + 9} \).
