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The problem involves a tangent-secant theorem with a circle. Here is the given information extracted from the image:

1. There is a circle with center \( O \) and a point \( Q \) outside the circle.
2. \( PQ \) is a tangent to the circle at point \( P \), and \( PR \) is a secant line that intersects the circle at \( R \) and \( Q \).
3. The length of \( PQ \) (the tangent) is given as \( 8 \) cm.
4. The length of \( PR \) (the secant segment from \( P \) to \( R \)) is given as \( 12 \) cm.
5. We need to find the radius \( r \) of the circle.

### Solving the Problem

The Tangent-Secant Theorem states that:

\[
PQ^2 = PR \cdot PQ
\]

Given that \( PQ = 8 \) cm and \( PR = 12 \) cm, substitute these values into the formula:

\[
8^2 = 12 \cdot PQ
\]

\[
64 = 12 \cdot PQ
\]

\[
PQ = \frac{64}{12}
\]

\[
PQ = \frac{16}{3} \approx 5.33 \text{ cm}
\]

Now, to find the radius \( r \) of the circle, we need to recognize that the length \( PR \) is composed of the sum of two parts: the external part (which is \( PQ = 8 \) cm) and the internal part (from \( P \) to the point of tangency \( Q \)). 

The radius can be derived by using the following relationship:

Using the formula:
\[
PQ^2 = PR \cdot PQ = 8^2 = (8 + x) \cdot x
\]
Where \( x \) is the distance of \( PQ \), hence:

\[
64 = (8 + x) \cdot x
\]

\[
x^2 + 8x - 64 = 0
\]

Solving the quadratic equation:

Using the quadratic formula:

\[
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

\[
x = \frac{-8 \pm \sqrt{8^2 - 4(1)(-64)}}{2(1)}
\]

\[
x = \frac{-8 \pm \sqrt{64 + 256}}{2}
\]

\[
x = \frac{-8 \pm \sqrt{320}}{2}
\]

\[
x = \frac{-8 \pm \sqrt{16 \cdot 20}}{2}
\]

\[
x = \frac{-8 \pm 4\sqrt{5}}{2}
\]

\[
x = -4 \pm 2\sqrt{5}
\]

Now since \( x \) must be positive

\[
x = -4 + 2\sqrt{5}
\]

Now using the value of x the radius of the circle is:

\[
r = \frac{16}{3}
\]

### Answer Choices

The answer choice is closest to \( r = \frac{16}{3} \approx 5.33 \text{ cm} \), and none of the options provided matches this value.

### Follow-up Questions and Tips

1. What is the Tangent-Secant theorem, and how is it applied in geometry?
2. Can you explain how the quadratic formula is derived?
3. What are some other practical applications of the tangent-secant theorem?
4. How does this problem illustrate the relationship between tangents and secants in a circle?
5. How would the problem change if the tangent-secant theorem wasn't applicable?

**Tip:** Always check your units and double-check calculations, especially when using formulas involving squares or roots.

A circle with center O has a tangent PQ and a secant PR. PQ = 8 cm and PR = 12 cm. Calculate the radius of the circle.

Learn how to calculate the radius of a circle given a tangent and a secant using the Tangent-Secant Theorem. This problem is suitable for Grades 9-11 and covers key geometry concepts and formulas.

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