Let d be the shortest and D be the longest distance between two circles `x = cos theta`, `y = sin theta` and `x = 3 + 2 cos theta, y = 3 + 2 sin theta`, then the value of (D – d) cannot be

Options

A. `6sqrt(2)`
B. 6
C. `3sqrt(2) + 2`
D. 5

Answer

To solve the problem, we need to determine the shortest (d) and longest (D) distances between the two circles defined by the parametric equations \( x = \cos \theta \), \( y = \sin \theta \) and \( x = 3 + 2 \cos \theta \), \( y = 3 + 2 \sin \theta \).

### Step-by-step Solution:

1. **Identify the Circles:**
- The first circle (C1) is defined by the equation \( x^2 + y^2 = 1 \). This circle has:
- Center \( O_1(0, 0) \)
- Radius \( r_1 = 1 \)
- The second circle (C2) is defined by the equation \( (x - 3)^2 + (y - 3)^2 = 4 \). This circle has:
- Center \( O_2(3, 3) \)
- Radius \( r_2 = 2 \)

2. **Calculate the Distance Between Centers:**
- The distance \( d_{O_1O_2} \) between the centers \( O_1(0, 0) \) and \( O_2(3, 3) \) can be calculated using the distance formula:
\[ d_{O_1O_2} = \sqrt{(3 - 0)^2 + (3 - 0)^2} = \sqrt{9 + 9} = 3\sqrt{2} \]

3. **Determine the Shortest Distance (d):**
- The shortest distance \( d \) between the two circles occurs when the circles are closest together:
\[ d = d_{O_1O_2} - (r_1 + r_2) = 3\sqrt{2} - (1 + 2) = 3\sqrt{2} - 3 \]

4. **Determine the Longest Distance (D):**
- The longest distance \( D \) between the two circles occurs when the circles are farthest apart:
\[ D = d_{O_1O_2} + (r_1 + r_2) = 3\sqrt{2} + (1 + 2) = 3\sqrt{2} + 3 \]

5. **Calculate \( D - d \):**
- Now, we calculate \( D - d \):
\[ D - d = (3\sqrt{2} + 3) - (3\sqrt{2} - 3) = 3 + 3 = 6 \]

6. **Conclusion:**
- The value of \( D - d \) is \( 6 \). The problem states that this value cannot be, which means the answer is \( 6 \).