A circle `S` of radius `' a '` is the director circle of another circle `S_1,S_1` is the director circle of circle `S_2` and so on. If the sum of the radii of all these circle is 2, then the value of `' a '` is (a) `2+sqrt(2)` (b) `2-1/(sqrt(2))` (c) `2-sqrt(2)` (d) `2+1/(sqrt(2))`

Answer

To solve the problem, we need to find the value of 'a' given that the sum of the radii of all the circles is 2. The circles are related through the property of director circles.

### Step-by-Step Solution:

1. **Understanding the Director Circle Property**:
The radius of the director circle of a circle with radius \( r \) is given by:
\[ R = \sqrt{2} \cdot r \]
This means that if \( S \) has radius \( a \), then the radius of circle \( S_1 \) will be:
\[ r_1 = \frac{a}{\sqrt{2}} \]

2. **Finding the Radius of Subsequent Circles**:
Continuing this process, the radius of circle \( S_2 \) will be:
\[ r_2 = \frac{r_1}{\sqrt{2}} = \frac{a}{\sqrt{2} \cdot \sqrt{2}} = \frac{a}{2} \]
The radius of circle \( S_3 \) will be:
\[ r_3 = \frac{r_2}{\sqrt{2}} = \frac{a}{2\sqrt{2}} = \frac{a}{2\sqrt{2}} \]
And so forth.

3. **Generalizing the Radius**:
The radius of the \( n \)-th circle can be expressed as:
\[ r_n = \frac{a}{(\sqrt{2})^n} \]

4. **Sum of the Radii**:
The total sum of the radii of all circles can be expressed as:
\[ S = a + \frac{a}{\sqrt{2}} + \frac{a}{2} + \frac{a}{2\sqrt{2}} + \ldots \]
This series can be factored out as:
\[ S = a \left( 1 + \frac{1}{\sqrt{2}} + \frac{1}{2} + \frac{1}{2\sqrt{2}} + \ldots \right) \]

5. **Identifying the Series**:
The series inside the parentheses is a geometric series with the first term \( 1 \) and a common ratio \( \frac{1}{\sqrt{2}} \). The sum of an infinite geometric series can be calculated using the formula:
\[ S = \frac{a}{1 - r} \]
where \( a \) is the first term and \( r \) is the common ratio. Here:
\[ S = \frac{1}{1 - \frac{1}{\sqrt{2}}} = \frac{\sqrt{2}}{\sqrt{2} - 1} \]

6. **Calculating the Total Sum**:
Thus, the total sum of the radii becomes:
\[ S = a \cdot \frac{\sqrt{2}}{\sqrt{2} - 1} \]

7. **Setting the Equation**:
We know that the total sum of the radii is equal to 2:
\[ a \cdot \frac{\sqrt{2}}{\sqrt{2} - 1} = 2 \]

8. **Solving for 'a'**:
Rearranging gives:
\[ a = 2 \cdot \frac{\sqrt{2} - 1}{\sqrt{2}} = 2 - \frac{2}{\sqrt{2}} = 2 - \sqrt{2} \]

9. **Final Answer**:
Therefore, the value of \( a \) is:
\[ a = 2 - \sqrt{2} \]

### Conclusion:
The correct option is (c) \( 2 - \sqrt{2} \).