https://file.aiquickdraw.com/mathbot/file/1727686238036v0rok0gn.jpg

To solve this problem, we can apply the *Power of a Point Theorem*. Specifically, we'll use the tangent-secant theorem, which states:

\[
PA^2 = PB \times PC
\]

### Given:
- \( PB = 4 \)
- \( BC = 8 \)
- Therefore, \( PC = PB + BC = 4 + 8 = 12 \)

### Applying the Theorem:
\[
PA^2 = PB \times PC = 4 \times 12 = 48
\]

Taking the square root of both sides:
\[
PA = \sqrt{48} = \sqrt{16 \times 3} = 4\sqrt{3}
\]

### Answer:
The length of \( PA \) is \( 4\sqrt{3} \), which corresponds to option **C**.

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Would you like more details or further clarification? Here are some related questions you might find useful:

1. What is the Power of a Point Theorem and how does it apply to circles?
2. How do you derive the tangent-secant theorem?
3. Can the Power of a Point Theorem be applied to other types of geometry problems?
4. What are some examples of secant-tangent relationships in real-world applications?
5. How does the relationship between the radius and tangent line of a circle help in solving geometric problems?

**Tip:** When dealing with circle geometry, always remember to explore the Power of a Point Theorem for secant and tangent combinations—it can save time and provide direct answers!

In the accompanying diagram, PA is tangent to circle O at A, PBC is a secant, PB = 4, and BC = 8. What is the length of PA?

Learn how to solve a geometric problem involving a tangent and secant of a circle, where PA is tangent, PB = 4, and BC = 8. The problem uses the Power of a Point Theorem and the Tangent-Secant Theorem to find the length of PA. Suitable for Grades 9-12.

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