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To solve this problem, we can use a property of tangents and secants. According to the Power of a Point Theorem, if a point outside a circle has a tangent and a secant line drawn to the circle, then the square of the length of the tangent segment is equal to the product of the entire secant length and its external part.

In the figure:
- \( TF \) is the tangent, so the length of the tangent \( TF = 15 \, \text{cm} \).
- \( FA \) is the external part of the secant line, and its length is \( FA = 9 \, \text{cm} \).
- \( FB \) is the entire secant, so \( FB = FA + AB = 2x + 3 \, \text{cm} \).

Now, we can apply the Power of a Point Theorem:
\[
TF^2 = FA \cdot FB
\]

Substitute the known values:
\[
15^2 = 9 \cdot (2x + 3)
\]
\[
225 = 9(2x + 3)
\]
\[
225 = 18x + 27
\]

Next, solve for \( x \):
\[
225 - 27 = 18x
\]
\[
198 = 18x
\]
\[
x = \frac{198}{18} = 11
\]

Thus, the value of \( x \) is \( \boxed{11} \).

Would you like more details or have any questions?

Here are five related questions:
1. What is the Power of a Point Theorem?
2. How do you find the external part of a secant?
3. What happens if both segments are tangents instead of a tangent and a secant?
4. Can you apply the same theorem if the point is inside the circle?
5. How would the formula change if the values of \( FA \) and \( FB \) were switched?

**Tip**: Remember that for tangent and secant problems, visualizing and labeling the diagram properly helps in applying the correct geometric theorems.

Find the value of x in the given figure where TF (tangent on point T) = 15 cm, FB = 2x + 3 cm, FA = 9 cm.

Solve a tangent and secant geometry problem by applying the Power of a Point Theorem. Given TF = 15 cm, FB = 2x + 3 cm, and FA = 9 cm, this step-by-step solution shows how to find the value of x.

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