`If A={x,y):x^(2)+y^(2)=25} and `
`B={(x,y):x^(2)+9y^(2)=144},` then ` A cap B ` contains

Options

A. one point
B. three point
C. two point
D. four point

Explanation

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

### Step 1: Define the equations
The set \( A \) is defined by the equation:
\[ x^2 + y^2 = 25 \quad \text{(Equation 1)} \]
This represents a circle with a radius of 5 centered at the origin.

The set \( B \) is defined by the equation:
\[ x^2 + 9y^2 = 144 \quad \text{(Equation 2)} \]
This represents an ellipse.

### Step 2: Solve for \( y^2 \) from Equation 1
From Equation 1, we can express \( y^2 \) in terms of \( x^2 \):
\[ y^2 = 25 - x^2 \quad \text{(Equation 3)} \]

### Step 3: Substitute Equation 3 into Equation 2
Now, substitute Equation 3 into Equation 2:
\[ x^2 + 9(25 - x^2) = 144 \]
Expanding this gives:
\[ x^2 + 225 - 9x^2 = 144 \]
Combining like terms results in:
\[ -8x^2 + 225 = 144 \]

### Step 4: Rearranging the equation
Rearranging the equation gives:
\[ -8x^2 = 144 - 225 \]
\[ -8x^2 = -81 \]
Dividing both sides by -8:
\[ x^2 = \frac{81}{8} \]

### Step 5: Solve for \( x \)
Taking the square root of both sides gives:
\[ x = \pm \sqrt{\frac{81}{8}} = \pm \frac{9}{\sqrt{8}} = \pm \frac{9\sqrt{2}}{4} \]

### Step 6: Substitute \( x \) back to find \( y \)
Now, substitute \( x^2 = \frac{81}{8} \) back into Equation 3 to find \( y^2 \):
\[ y^2 = 25 - \frac{81}{8} \]
Converting 25 to a fraction with a denominator of 8:
\[ y^2 = \frac{200}{8} - \frac{81}{8} = \frac{119}{8} \]

### Step 7: Solve for \( y \)
Taking the square root of both sides gives:
\[ y = \pm \sqrt{\frac{119}{8}} = \pm \frac{\sqrt{119}}{2\sqrt{2}} = \pm \frac{\sqrt{119}}{4} \]

### Step 8: Find the intersection points
The intersection points \( A \cap B \) are:
1. \( \left( \frac{9\sqrt{2}}{4}, \frac{\sqrt{119}}{4} \right) \)
2. \( \left( \frac{9\sqrt{2}}{4}, -\frac{\sqrt{119}}{4} \right) \)
3. \( \left( -\frac{9\sqrt{2}}{4}, \frac{\sqrt{119}}{4} \right) \)
4. \( \left( -\frac{9\sqrt{2}}{4}, -\frac{\sqrt{119}}{4} \right) \)

Thus, \( A \cap B \) contains 4 points.