Let PQ and RS be tangents at the extremities of the diameter PR of a circle of radius r. If PS and RQ intersect at a point X on the circumference of the circle, then 2r equals

Options

A. `sqrt( PQ. RS )`
B. `( PQ+RS)/( 2)`
C. `(2PQ.RS)/( PQ+RS)`
D. `sqrt((PQ^(2)+RS^(2))/(2))`

Answer

To solve the problem, we need to establish the relationship between the tangents PQ and RS, the diameter PR, and the point X where the lines PS and RQ intersect. We will use properties of tangents and the right triangle formed by the diameter and the tangents.

### Step-by-Step Solution:

1. **Understanding the Setup**:
- Let the center of the circle be O, and the radius of the circle be r.
- The points P and R are the endpoints of the diameter PR, and Q and S are points where the tangents PQ and RS touch the circle.
- The tangents PQ and RS intersect at point X on the circumference of the circle.

2. **Using the Right Triangle Property**:
- Since PR is a diameter, by the property of circles, any angle subtended by the diameter at the circumference (angle PXQ) is a right angle (90 degrees).
- Therefore, we can apply trigonometric identities to the right triangle formed.

3. **Applying Trigonometric Ratios**:
- From triangle PXQ, we can write:
\[ \tan(\theta) = \frac{PQ}{OX} \quad \text{(where OX = r)} \]
- Thus, we have:
\[ \tan(\theta) = \frac{PQ}{2r} \quad \text{(since OX = r and PX = r)} \]
- Similarly, for triangle RXQ:
\[ \tan(90^\circ - \theta) = \frac{RS}{OX} = \frac{RS}{2r} \]
- This implies:
\[ \cot(\theta) = \frac{RS}{2r} \]

4. **Setting Up the Equations**:
- We now have two equations:
1. \( \tan(\theta) = \frac{PQ}{2r} \) (Equation 1)
2. \( \cot(\theta) = \frac{RS}{2r} \) (Equation 2)

5. **Multiplying the Equations**:
- Multiply Equation 1 and Equation 2:
\[ \tan(\theta) \cdot \cot(\theta) = \frac{PQ}{2r} \cdot \frac{RS}{2r} \]
- Since \( \tan(\theta) \cdot \cot(\theta) = 1 \), we have:
\[ 1 = \frac{PQ \cdot RS}{4r^2} \]

6. **Solving for 2r**:
- Rearranging the equation gives us:
\[ 4r^2 = PQ \cdot RS \]
- Taking the square root of both sides:
\[ 2r = \sqrt{PQ \cdot RS} \]

### Conclusion:
Thus, we find that:
\[ 2r = \sqrt{PQ \cdot RS} \]