Three circles `C_(1),C_(2),C_(3)` with radii `r_(1),r_(2),r_(3)(r_(1)ltr_(2)ltr_(3))` respectively are given as `r_(1)=2`, and `r_(3)=8` they are placed such that `C_(2)` lines to the right of `C_(1)` and touches it externally `C_(3)` lies ot the right of `C_(2)` and touches it externally. There exist two stratight lines each of whic is a direct common tangent simultaneously to all the three circles then `r_(2)` is equal to

Options

A. `r_(2)=4`
B. `r_(2)=5`
C. `r_(2)=10`
D. `r_(2)=16`

Explanation

To solve the problem, we need to find the radius \( r_2 \) of the circle \( C_2 \) given that \( r_1 = 2 \) and \( r_3 = 8 \). The circles are arranged such that \( C_1 \) and \( C_2 \) touch each other externally, and \( C_2 \) and \( C_3 \) also touch each other externally. Additionally, there are two direct common tangents to all three circles.

### Step-by-step Solution:

1. **Identify the given values:**
- \( r_1 = 2 \)
- \( r_3 = 8 \)
- We need to find \( r_2 \).

2. **Understand the relationship between the circles:**
- The distance between the centers of \( C_1 \) and \( C_2 \) is \( r_1 + r_2 \).
- The distance between the centers of \( C_2 \) and \( C_3 \) is \( r_2 + r_3 \).

3. **Use the formula for the length of the direct common tangent:**
- The formula for the length of the direct common tangent \( D \) between two circles with radii \( r_1 \) and \( r_2 \) is given by:
\[ D = \sqrt{(r_1 + r_2)^2 - (r_1 - r_2)^2} \]
- Simplifying this gives:
\[ D = \sqrt{4r_1 r_2} \]

4. **Set up the equations for the two pairs of circles:**
- For circles \( C_1 \) and \( C_2 \):
\[ D_{12} = \sqrt{4r_1 r_2} \]
- For circles \( C_2 \) and \( C_3 \):
\[ D_{23} = \sqrt{4r_2 r_3} \]

5. **Set the distances equal to each other:**
- Since both \( D_{12} \) and \( D_{23} \) are equal to the same tangent length \( D \):
\[ \frac{r_1 + r_2}{\sqrt{4r_1 r_2}} = \frac{r_2 + r_3}{\sqrt{4r_2 r_3}} \]

6. **Cross-multiply and simplify:**
- Cross-multiplying gives:
\[ (r_1 + r_2) \sqrt{4r_2 r_3} = (r_2 + r_3) \sqrt{4r_1 r_2} \]
- Squaring both sides leads to:
\[ (r_1 + r_2)^2 \cdot 4r_2 r_3 = (r_2 + r_3)^2 \cdot 4r_1 r_2 \]

7. **Substituting known values:**
- Substitute \( r_1 = 2 \) and \( r_3 = 8 \):
\[ (2 + r_2)^2 \cdot 4r_2 \cdot 8 = (r_2 + 8)^2 \cdot 4 \cdot 2 \cdot r_2 \]

8. **Cancel out common factors and simplify:**
- Dividing both sides by \( 4r_2 \):
\[ (2 + r_2)^2 \cdot 8 = (r_2 + 8)^2 \cdot 2 \]
- Simplifying further leads to:
\[ 8(2 + r_2)^2 = 2(r_2 + 8)^2 \]

9. **Expand and solve for \( r_2 \):**
- Expanding both sides:
\[ 8(4 + 4r_2 + r_2^2) = 2(r_2^2 + 16r_2 + 64) \]
- This simplifies to:
\[ 32 + 32r_2 + 8r_2^2 = 2r_2^2 + 32r_2 + 128 \]
- Rearranging gives:
\[ 6r_2^2 - 96 = 0 \]
- Solving for \( r_2 \):
\[ r_2^2 = 16 \implies r_2 = 4 \]

### Final Answer:
Thus, the radius \( r_2 \) of circle \( C_2 \) is \( \boxed{4} \).