There are two circles `C_(1)` and `C_(2)` touching each other and the coordinate axes, if `C_(1)` is smaller than `C_(2)` and its radius is 2 units, then radius of `C_(2)`, is

Options

A. `6+4sqrt(2)`
B. `2+2sqrt(2)`
C. `3+2sqrt(2)`
D. none of these

Explanation

To solve the problem, we need to find the radius of the larger circle \( C_2 \) given that the radius of the smaller circle \( C_1 \) is 2 units. Both circles touch each other and the coordinate axes.

### Step-by-Step Solution:

1. **Understanding the Configuration**:
- Let the radius of circle \( C_1 \) be \( r_1 = 2 \) units.
- Let the radius of circle \( C_2 \) be \( r_2 \) (which we need to find).
- Both circles touch each other and the coordinate axes.

2. **Positioning the Circles**:
- Circle \( C_1 \) can be positioned with its center at \( (2, 2) \) since it touches both the x-axis and y-axis.
- Circle \( C_2 \) will have its center at \( (r_2, r_2) \) since it also touches both axes.

3. **Distance Between the Centers**:
- The distance between the centers of the two circles \( C_1 \) and \( C_2 \) can be calculated using the distance formula:
\[ \text{Distance} = \sqrt{(r_2 - 2)^2 + (r_2 - 2)^2} = \sqrt{2(r_2 - 2)^2} = \sqrt{2} |r_2 - 2| \]

4. **Condition for Touching Circles**:
- For the circles to touch each other, the distance between their centers must equal the sum of their radii:
\[ \sqrt{2} |r_2 - 2| = r_1 + r_2 = 2 + r_2 \]
- This simplifies to:
\[ \sqrt{2} |r_2 - 2| = 2 + r_2 \]

5. **Solving the Equation**:
- We can consider two cases for \( |r_2 - 2| \):
- Case 1: \( r_2 - 2 \geq 0 \) (i.e., \( r_2 \geq 2 \)):
\[ \sqrt{2}(r_2 - 2) = 2 + r_2 \]
\[ \sqrt{2}r_2 - 2\sqrt{2} = 2 + r_2 \]
\[ (\sqrt{2} - 1)r_2 = 2 + 2\sqrt{2} \]
\[ r_2 = \frac{2 + 2\sqrt{2}}{\sqrt{2} - 1} \]

- Case 2: \( r_2 - 2 < 0 \) (i.e., \( r_2 < 2 \)):
\[ \sqrt{2}(2 - r_2) = 2 + r_2 \]
\[ 2\sqrt{2} - \sqrt{2}r_2 = 2 + r_2 \]
\[ (1 + \sqrt{2})r_2 = 2\sqrt{2} - 2 \]
\[ r_2 = \frac{2(\sqrt{2} - 1)}{1 + \sqrt{2}} \]

6. **Calculating the Radius**:
- We will only consider the first case since \( C_2 \) is larger than \( C_1 \):
\[ r_2 = \frac{2 + 2\sqrt{2}}{\sqrt{2} - 1} \]
- Rationalizing the denominator:
\[ r_2 = \frac{(2 + 2\sqrt{2})(\sqrt{2} + 1)}{(\sqrt{2} - 1)(\sqrt{2} + 1)} = \frac{(2\sqrt{2} + 2 + 4 + 2\sqrt{2})}{2 - 1} = 6 + 4\sqrt{2} \]

### Final Answer:
The radius of circle \( C_2 \) is \( 6 + 4\sqrt{2} \) units.

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