O and O' are the centres of two circles which touch each other externally at P. AB is a common tangent. Find `angleAPO`

Options

A. `90^(@)`
B. `120^(@)`
C. `60^(@)`
D. data insufficient

Answer

To solve the problem, we need to find the angle \( \angle APO \) where \( O \) and \( O' \) are the centers of two circles that touch each other externally at point \( P \), and \( AB \) is a common tangent to both circles.

### Step-by-Step Solution:

1. **Understand the Configuration**:
- We have two circles touching each other externally at point \( P \).
- Let \( O \) be the center of the first circle and \( O' \) be the center of the second circle.
- The tangent line \( AB \) touches both circles at point \( P \).

2. **Draw the Diagram**:
- Draw two circles touching at point \( P \).
- Mark the centers \( O \) and \( O' \).
- Draw the tangent line \( AB \) that touches the circles at point \( P \).

3. **Identify Key Points**:
- The points of tangency create right angles with the radius at the point of tangency. Therefore, \( OP \) is perpendicular to \( AB \) at point \( P \).

4. **Analyze the Angles**:
- Since \( OP \) is perpendicular to the tangent line \( AB \), we have:
\[ \angle OPA = 90^\circ \]
- The triangle \( OAP \) consists of two sides \( OP \) and \( OA \) (where \( OA \) is the line connecting the center \( O \) to point \( A \) on the tangent).

5. **Use the Properties of Isosceles Triangle**:
- The angles \( \angle OAP \) and \( \angle OPA \) are equal because \( OP \) and \( O'A \) are both radii of their respective circles.
- Let \( \angle OAP = \angle OPA = x \).

6. **Set Up the Equation**:
- In triangle \( OAP \):
\[ x + x + 90^\circ = 180^\circ \]
- This simplifies to:
\[ 2x + 90^\circ = 180^\circ \]
- Subtract \( 90^\circ \) from both sides:
\[ 2x = 90^\circ \]
- Divide by 2:
\[ x = 45^\circ \]

7. **Conclusion**:
- Thus, \( \angle OAP = 45^\circ \).
- Therefore, \( \angle APO = 90^\circ - \angle OAP = 90^\circ - 45^\circ = 45^\circ \).

### Final Answer:
The angle \( \angle APO \) is \( 45^\circ \).