Two circles of radii `r_(1)` and `r_(2), r_(1) gt r_(2) ge2` touch each other externally. If `theta` be the angle between the direct common tangents, then,

Options

A. `theta=sin^(-1)((r_1+r_2)/(r_1-r_2))`
B. `theta=2sin^(-1)((r_1-r_2)/(r_1+r_2))`
C. `theta=2cos^(-1)((2sqrt(r_1r_2))/(r_1+r_2))`
D. none of these

Answer

To solve the problem, we need to find the angle \( \theta \) between the direct common tangents of two circles with radii \( r_1 \) and \( r_2 \) (where \( r_1 > r_2 \geq 2 \)) that touch each other externally.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- We have two circles that touch each other externally. The centers of the circles are separated by a distance equal to the sum of their radii, \( r_1 + r_2 \).

2. **Using Triangle Properties**:
- Let the centers of the circles be \( O_1 \) and \( O_2 \). The direct common tangents touch the circles at points \( A \) and \( B \).
- The line segment \( O_1O_2 \) forms a triangle with the tangents \( O_1A \) and \( O_2B \).

3. **Identifying Angles**:
- Let \( \alpha \) be the angle between the line connecting the centers \( O_1O_2 \) and one of the tangents, say \( O_1A \).
- The angle \( \theta \) between the two direct common tangents can be expressed in terms of \( \alpha \) as \( \theta = 2\alpha \).

4. **Using Sine Rule**:
- In triangle \( O_1AE \) (where \( E \) is the point where the tangent meets the line joining the centers), we can use the sine rule:
\[ \sin(\alpha) = \frac{DE}{EC} \]
- Here, \( DE = r_2 \) and \( EC \) can be expressed in terms of \( r_1 \) and \( r_2 \).

5. **Setting Up the Equation**:
- From the triangle similarity, we can derive:
\[ \frac{r_1}{r_2} = \frac{DE + EC}{EC} \]
- Rearranging gives:
\[ \frac{r_1}{r_2} - 1 = \frac{DE}{EC} \]

6. **Finding \( \sin(\alpha) \)**:
- We can express \( EC \) in terms of \( r_1 \) and \( r_2 \):
\[ EC = \frac{r_1 + r_2}{r_2} \cdot \sin(\alpha) \]
- Substituting back, we can find:
\[ \sin(\alpha) = \frac{r_1 - r_2}{r_1 + r_2} \]

7. **Finding \( \theta \)**:
- Since \( \theta = 2\alpha \), we can find \( \theta \) using the inverse sine function:
\[ \alpha = \sin^{-1}\left(\frac{r_1 - r_2}{r_1 + r_2}\right) \]
- Therefore, the angle \( \theta \) is:
\[ \theta = 2 \sin^{-1}\left(\frac{r_1 - r_2}{r_1 + r_2}\right) \]

### Final Answer:
\[ \theta = 2 \sin^{-1}\left(\frac{r_1 - r_2}{r_1 + r_2}\right) \]