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

Options

A. `A cap B=phi`
B. `Acap B` is singlcton set
C. `Acap B ` contains two elements only
D. `A cap B ` contains three elements only .

Answer

To solve the problem, we need to find the intersection of the two sets A and B defined as follows:

1. **Set A**: \( A = \{(x, y) : x^2 + y^2 = 4, \, x, y \in \mathbb{R}\} \)
- This represents a circle with a radius of 2 centered at the origin (0,0).

2. **Set B**: \( B = \{(x, y) : y = |x|, \, x, y \in \mathbb{R}\} \)
- This represents the V-shaped graph of the absolute value function, which consists of two lines: \( y = x \) for \( x \geq 0 \) and \( y = -x \) for \( x < 0 \).

Now, we need to find the points that are common to both sets, i.e., \( A \cap B \).

### Step 1: Substitute \( y = |x| \) into the equation of the circle

The equation of the circle is:
\[ x^2 + y^2 = 4 \]
Substituting \( y = |x| \):
- For \( x \geq 0 \), \( y = x \):
\[ x^2 + x^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2} \]
Then, \( y = \sqrt{2} \). So, one point is \( (\sqrt{2}, \sqrt{2}) \).

- For \( x < 0 \), \( y = -x \):
\[ x^2 + (-x)^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = -\sqrt{2} \]
Then, \( y = -(-\sqrt{2}) = \sqrt{2} \). So, another point is \( (-\sqrt{2}, \sqrt{2}) \).

### Step 2: List the points found

The points found in the intersection \( A \cap B \) are:
1. \( (\sqrt{2}, \sqrt{2}) \)
2. \( (-\sqrt{2}, \sqrt{2}) \)

### Step 3: Count the number of elements in the intersection

Thus, the number of elements in \( A \cap B \) is 2.

### Final Answer
The number of elements in \( A \cap B \) is **2**.

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