`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 \( B \).

### Step 1: Define the sets
Set \( A \) is defined by the equation \( x^2 + y^2 = 4 \). This represents a circle centered at the origin (0,0) with a radius of 2.

Set \( B \) is defined by the equation \( y = |x| \). This represents a V-shaped graph that opens upwards, consisting of two lines: \( y = x \) for \( x \geq 0 \) and \( y = -x \) for \( x < 0 \).

### Step 2: Graph the sets
To visualize the problem, we can sketch both graphs:
- The circle from set \( A \) will intersect the y-axis at (0,2) and (0,-2) and the x-axis at (2,0) and (-2,0).
- The V-shaped graph from set \( B \) will intersect the y-axis at (0,0) and will have lines extending outwards from this point.

### Step 3: Find the intersection points
To find the intersection points, we need to solve the equations simultaneously.

1. Substitute \( y = |x| \) into the circle's equation \( x^2 + y^2 = 4 \).
- For \( y = x \) (when \( x \geq 0 \)):
\[ x^2 + x^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = \sqrt{2} \text{ (since } x \geq 0\text{)} \]
Thus, \( y = \sqrt{2} \). So one intersection point is \( \left(\sqrt{2}, \sqrt{2}\right) \).

- For \( y = -x \) (when \( x < 0 \)):
\[ x^2 + (-x)^2 = 4 \implies 2x^2 = 4 \implies x^2 = 2 \implies x = -\sqrt{2} \text{ (since } x < 0\text{)} \]
Thus, \( y = -\sqrt{2} \). So the other intersection point is \( \left(-\sqrt{2}, -\sqrt{2}\right) \).

### Step 4: Conclusion
The intersection \( A \cap B \) contains the points \( \left(\sqrt{2}, \sqrt{2}\right) \) and \( \left(-\sqrt{2}, -\sqrt{2}\right) \). Therefore, the intersection of sets \( A \) and \( B \) contains 2 elements.

### Final Answer:
The intersection \( A \cap B = \left\{ \left(\sqrt{2}, \sqrt{2}\right), \left(-\sqrt{2}, -\sqrt{2}\right) \right\} \).

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