P is a point outside a circle and is 26cm away from its centre.A secant PAB drawn from intersects the circle at points A and B such that PB = 32 cm and PA = 18 cm. The radius of the circle ( in cm ) is `:`

Options

A. 12
B. 8
C. 10
D. 13

Answer

To find the radius of the circle, we can use the Power of a Point theorem, which states that if a point P is outside a circle, then the product of the lengths of the segments of any secant line drawn from P to the circle is equal to the square of the length of the tangent segment from P to the circle.

Let's denote:
- O = center of the circle
- R = radius of the circle
- OP = distance from point P to the center O = 26 cm
- PA = length from P to point A = 18 cm
- PB = length from P to point B = 32 cm

### Step-by-step solution:

1. **Identify the segments**:
- We know that PA = 18 cm and PB = 32 cm.
- The segment lengths can be expressed as:
- \( PB = PA + AB \)
- Therefore, \( AB = PB - PA = 32 - 18 = 14 \) cm.

2. **Using the Power of a Point theorem**:
- According to the theorem, we have:
\[ PA \cdot PB = PO^2 - R^2 \]
- Substituting the values:
\[ 18 \cdot 32 = (26)^2 - R^2 \]

3. **Calculate the left side**:
- Calculate \( 18 \cdot 32 \):
\[ 18 \cdot 32 = 576 \]

4. **Calculate the right side**:
- Calculate \( (26)^2 \):
\[ (26)^2 = 676 \]
- Now, substituting back into the equation:
\[ 576 = 676 - R^2 \]

5. **Rearranging the equation**:
- Rearranging gives:
\[ R^2 = 676 - 576 \]
- Calculate \( 676 - 576 \):
\[ R^2 = 100 \]

6. **Finding the radius**:
- Taking the square root of both sides:
\[ R = \sqrt{100} = 10 \text{ cm} \]

### Final Answer:
The radius of the circle is \( R = 10 \) cm.