`if A={(x,y):y=(4)/(x),xne0}and `
` B= {(x,y):x^(2)+y^(2)=8,x,yin R}`,then

Options

A. `A cap B=phi`
B. `Acap B ` contains one point only
C. `A cap B `contains two points only
D. `A cap B ` is a multiple of 4 points only .

Answer

To find the intersection of the sets \( A \) and \( B \), we need to solve the equations that define these sets.

1. **Define the sets**:
- Set \( A = \{(x, y) : y = \frac{4}{x}, x \neq 0\} \)
- Set \( B = \{(x, y) : x^2 + y^2 = 8, x, y \in \mathbb{R}\} \)

2. **Substitute the equation of set \( A \) into the equation of set \( B \)**:
- Since \( y = \frac{4}{x} \), substitute this into the equation of set \( B \):
\[ x^2 + \left(\frac{4}{x}\right)^2 = 8 \]

3. **Simplify the equation**:
- This becomes:
\[ x^2 + \frac{16}{x^2} = 8 \]

4. **Multiply through by \( x^2 \) to eliminate the fraction** (note that \( x \neq 0 \)):
\[ x^4 - 8x^2 + 16 = 0 \]

5. **Let \( z = x^2 \)** (this substitution simplifies the equation):
- The equation now becomes:
\[ z^2 - 8z + 16 = 0 \]

6. **Factor the quadratic equation**:
- This can be factored as:
\[ (z - 4)^2 = 0 \]
- Thus, \( z = 4 \).

7. **Solve for \( x \)**:
- Since \( z = x^2 \), we have:
\[ x^2 = 4 \implies x = 2 \text{ or } x = -2 \]

8. **Find corresponding \( y \) values**:
- For \( x = 2 \):
\[ y = \frac{4}{2} = 2 \]
- For \( x = -2 \):
\[ y = \frac{4}{-2} = -2 \]

9. **List the intersection points**:
- The points in the intersection \( A \cap B \) are:
\[ (2, 2) \text{ and } (-2, -2) \]

10. **Conclusion**:
- Therefore, the number of elements in the intersection \( A \cap B \) is \( 2 \).