Which of the following expresses the circumference of a circle inscribed in a sector `O A B` with radius `Ra n dA B=2a ?` `2pi(R a)/(R+a)` (b) `(2piR^2)/a` `2pi(r-a)^2` (d) `2piR/(R-a)`

Options

A. `2pi (Ra)/(R + a)`
B. `(2pi R^(2))/(a)`
C. `2pi (R -a)^(2)`
D. `2pi(R)/(R -a)`

Explanation

To solve the problem of finding the circumference of a circle inscribed in a sector OAB with radius \( R \) and \( AB = 2a \), we can follow these steps:

### Step-by-Step Solution:

1. **Understand the Geometry**:
- We have a sector OAB with a radius \( R \) and the length of the chord \( AB = 2a \).
- The inscribed circle touches the two radii and the chord AB.

2. **Identify the Relationship**:
- The radius \( r \) of the inscribed circle can be related to the radius \( R \) of the sector and the length of the chord \( AB \).
- The relationship can be derived using the sine rule in the triangle formed by the center O, point A, and point B.

3. **Use the Sine Rule**:
- From the triangle OAB, we can write:
\[ \sin \theta = \frac{R}{OH} = \frac{A}{OA} \]
- Here, \( OH \) is the distance from the center O to the chord AB, and \( A \) is half the length of the chord, which is \( a \).

4. **Set Up the Equation**:
- From the sine rule, we can express:
\[ \frac{R}{A} = \frac{R - r}{R} \]
- Rearranging gives us:
\[ \frac{R}{A} + \frac{r}{R} = 1 \]

5. **Find the Value of r**:
- By taking the least common multiple (LCM) of the fractions, we can rewrite the equation:
\[ rA + rR = A \cdot R \]
- This simplifies to:
\[ r = \frac{rA}{R + A} \]

6. **Circumference of the Inscribed Circle**:
- The circumference \( C \) of the inscribed circle is given by the formula:
\[ C = 2\pi r \]
- Substituting the value of \( r \) we found:
\[ C = 2\pi \left(\frac{rA}{R + A}\right) \]
- This simplifies to:
\[ C = \frac{2\pi rA}{R + A} \]

7. **Conclusion**:
- Therefore, the expression for the circumference of the circle inscribed in the sector OAB is:
\[ C = \frac{2\pi rA}{R + A} \]
- This matches with option (a): \( \frac{2\pi R a}{R + a} \).

### Final Answer:
The correct option is (a) \( \frac{2\pi R a}{R + a} \).