A tangent PQ at point P of a circle of radius 12 cm meets a line through the centre O to a point Q so that OQ = 20 cm . Length PQ is :

Options

A. 14
B. 15
C. 16
D. 10

Answer

To find the length of the tangent \( PQ \) from point \( P \) on the circle to point \( Q \) on the line through the center \( O \), we can use the properties of right triangles and the Pythagorean theorem.

### Step-by-Step Solution:

1. **Identify the Given Information:**
- Radius of the circle \( r = 12 \, \text{cm} \)
- Length of line segment \( OQ = 20 \, \text{cm} \)

2. **Understand the Geometry:**
- The tangent \( PQ \) at point \( P \) is perpendicular to the radius \( OP \) at point \( P \).
- This forms a right triangle \( OPQ \) where:
- \( OP \) is the radius of the circle.
- \( OQ \) is the distance from the center \( O \) to point \( Q \).
- \( PQ \) is the length of the tangent we need to find.

3. **Apply the Pythagorean Theorem:**
- In triangle \( OPQ \):
\[ OQ^2 = OP^2 + PQ^2 \]

4. **Substitute the Known Values:**
- We know \( OQ = 20 \, \text{cm} \) and \( OP = 12 \, \text{cm} \).
- Substitute these values into the equation:
\[ 20^2 = 12^2 + PQ^2 \]
\[ 400 = 144 + PQ^2 \]

5. **Solve for \( PQ^2 \):**
- Rearranging the equation gives:
\[ PQ^2 = 400 - 144 \]
\[ PQ^2 = 256 \]

6. **Find \( PQ \):**
- Taking the square root of both sides:
\[ PQ = \sqrt{256} = 16 \, \text{cm} \]

### Final Answer:
The length of \( PQ \) is \( 16 \, \text{cm} \).