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

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

### Step 1: Substitute the equation of Set A into Set B

We know that \( y = \frac{4}{x} \) from Set A. We will substitute this expression for \( y \) into the equation of Set B.

\[ x^2 + \left(\frac{4}{x}\right)^2 = 8 \]

### Step 2: Simplify the equation

Now, we simplify the equation:

\[ x^2 + \frac{16}{x^2} = 8 \]

### Step 3: Multiply through by \( x^2 \) to eliminate the fraction

To eliminate the fraction, multiply the entire equation by \( x^2 \) (noting that \( x \neq 0 \)):

\[ x^4 + 16 = 8x^2 \]

### Step 4: Rearrange the equation

Rearranging gives us:

\[ x^4 - 8x^2 + 16 = 0 \]

### Step 5: Let \( z = x^2 \)

Let \( z = x^2 \). The equation becomes:

\[ z^2 - 8z + 16 = 0 \]

### Step 6: Factor the quadratic equation

This can be factored as:

\[ (z - 4)^2 = 0 \]

### Step 7: Solve for \( z \)

Setting the factor equal to zero gives:

\[ z - 4 = 0 \implies z = 4 \]

### Step 8: Substitute back to find \( x \)

Since \( z = x^2 \):

\[ x^2 = 4 \implies x = 2 \text{ or } x = -2 \]

### Step 9: Find corresponding \( y \) values

Now we substitute \( x \) back into the equation for \( y \):

For \( x = 2 \):
\[ y = \frac{4}{2} = 2 \]

For \( x = -2 \):
\[ y = \frac{4}{-2} = -2 \]

### Step 10: Write the intersection points

Thus, the points of intersection \( A \cap B \) are:

\[ (2, 2) \text{ and } (-2, -2) \]

### Final Result

The intersection of sets A and B contains the points \( (2, 2) \) and \( (-2, -2) \).

---