P is a point outside the circle at a distance of `6 . 5` cm from centre O of the circle . PR be a secant such that it intersects the circle at Q and R . If `PQ = 4.5 ` cm and `QR = 3.5` cm , then what is the radius ( in cm ) of the circle ?

Options

A. `1 .5`
B. `2 `
C. `2. 5`
D. `3 `

Answer

To find the radius of the circle given the distances and segments, we can use the secant-tangent theorem. Here’s how to solve the problem step by step:

### Step-by-Step Solution

1. **Identify the Given Values**:
- Distance from point P to the center O of the circle: \( PO = 6.5 \) cm.
- Length of segment PQ: \( PQ = 4.5 \) cm.
- Length of segment QR: \( QR = 3.5 \) cm.

2. **Calculate the Length of PR**:
- Since \( PR = PQ + QR \):
\[ PR = 4.5 \, \text{cm} + 3.5 \, \text{cm} = 8 \, \text{cm}. \]

3. **Set Up the Secant-Tangent Theorem**:
- According to the secant-tangent theorem, we have:
\[ PQ \times PR = PO^2 - r^2, \]
where \( r \) is the radius of the circle.

4. **Substitute the Known Values**:
- Substitute \( PQ = 4.5 \) cm, \( PR = 8 \) cm, and \( PO = 6.5 \) cm into the equation:
\[ 4.5 \times 8 = 6.5^2 - r^2. \]

5. **Calculate the Left Side**:
- Calculate \( 4.5 \times 8 \):
\[ 4.5 \times 8 = 36. \]

6. **Calculate \( 6.5^2 \)**:
- Calculate \( 6.5^2 \):
\[ 6.5^2 = 42.25. \]

7. **Set Up the Equation**:
- Now we have:
\[ 36 = 42.25 - r^2. \]

8. **Rearranging the Equation**:
- Rearranging gives:
\[ r^2 = 42.25 - 36. \]
- Simplifying:
\[ r^2 = 6.25. \]

9. **Taking the Square Root**:
- Taking the square root of both sides gives:
\[ r = \sqrt{6.25} = 2.5 \, \text{cm}. \]

### Conclusion
The radius of the circle is \( 2.5 \) cm.