`if A={(x,y):x^(2)+y^(2)=4,x,y in R } and`
`B={(x,y): x^(2)+y^(2)= 9,x,yin R}`,then

Options

A. `A-B= phi `
B. `B-A=B`
C. `A cap B ne phi `
D. `A cap B=A`

Answer

To solve the problem, we need to analyze the sets A and B defined in the question.

### Step 1: Define the sets A and B
The set A is defined as:
\[ A = \{(x,y) : x^2 + y^2 = 4, \, x, y \in \mathbb{R}\} \]
This represents a circle with a radius of 2 centered at the origin (0,0).

The set B is defined as:
\[ B = \{(x,y) : x^2 + y^2 = 9, \, x, y \in \mathbb{R}\} \]
This represents a circle with a radius of 3 centered at the origin (0,0).

### Step 2: Find A ∩ B (Intersection of A and B)
To find the intersection of sets A and B, we need to determine if there are any points (x, y) that satisfy both equations:
1. \( x^2 + y^2 = 4 \) (from set A)
2. \( x^2 + y^2 = 9 \) (from set B)

Since the left-hand sides of both equations cannot be equal (4 cannot equal 9), there are no points that satisfy both conditions. Therefore:
\[ A \cap B = \emptyset \]
This means that the intersection of sets A and B is the empty set.

### Step 3: Find A - B (Difference of A and B)
The difference \( A - B \) is defined as the set of elements that are in A but not in B. Since there are no common points between A and B, we have:
\[ A - B = A \]
Thus, \( A - B \) is simply the set A itself.

### Step 4: Find B - A (Difference of B and A)
Similarly, the difference \( B - A \) is defined as the set of elements that are in B but not in A. Again, since there are no common points between A and B, we have:
\[ B - A = B \]
Thus, \( B - A \) is simply the set B itself.

### Summary of Results
- \( A \cap B = \emptyset \)
- \( A - B = A \)
- \( B - A = B \)