The tangents are drawn at the extremities of diameter AB of a circle with centre P. If a tangent to the circle at the point C Intersects the other two tangents at Q and R, then the measure of the `/_QPR` is

Options

A. `45^@`
B. `60^@`
C. `90^@`
D. `180^@`

Answer

To solve the problem, we need to find the measure of the angle \( \angle QPR \) formed by the tangents at points Q and R, which intersect at point P, where the circle has its center.

### Step-by-Step Solution:

1. **Draw the Circle and Diameter**:
- Begin by drawing a circle with center P. Mark the diameter AB on the circle.

2. **Draw Tangents at A and B**:
- From points A and B, draw tangents to the circle. Let these tangents intersect at points Q and R.

3. **Identify Point C on the Circle**:
- Choose a point C on the circumference of the circle. This point will be where the tangent intersects the circle.

4. **Understanding the Angles**:
- Since the tangents at points A and B are perpendicular to the radius at those points, we know that \( \angle OAP = 90^\circ \) and \( \angle OBP = 90^\circ \).

5. **Using the Properties of Tangents**:
- The angle between two tangents drawn from an external point (in this case, point C) to a circle is equal to half the difference of the angles subtended by the points of tangency at the center of the circle.

6. **Calculate Angles**:
- Since the tangents at A and B are perpendicular to the radius, we can conclude that \( \angle BRQ = 90^\circ \) and \( \angle PKR = 90^\circ \).

7. **Using Triangle Angle Sum Property**:
- In triangle PQR, the sum of angles is \( 180^\circ \). Therefore, we can write:
\[ \angle PRQ + \angle PKR + \angle QPR = 180^\circ \]
- Given that \( \angle PRQ = 45^\circ \) and \( \angle PKR = 45^\circ \):
\[ 45^\circ + 45^\circ + \angle QPR = 180^\circ \]

8. **Solve for \( \angle QPR \)**:
- Simplifying the equation:
\[ 90^\circ + \angle QPR = 180^\circ \]
- Therefore:
\[ \angle QPR = 180^\circ - 90^\circ = 90^\circ \]

### Final Answer:
The measure of \( \angle QPR \) is \( 90^\circ \).