`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 analyze the equations given for each set.

1. **Identify the equations of the sets:**
- Set \( A \): \( x^2 + y^2 = 25 \)
- Set \( B \): \( x^2 + 9y^2 = 144 \)

2. **Understand the shapes represented by the equations:**
- The equation \( x^2 + y^2 = 25 \) represents a circle centered at the origin (0, 0) with a radius of \( 5 \).
- The equation \( x^2 + 9y^2 = 144 \) represents an ellipse centered at the origin. To put it in standard form, we can divide the entire equation by \( 144 \):
\[ \frac{x^2}{144} + \frac{y^2}{16} = 1 \]
This indicates that the semi-major axis (along the x-axis) is \( 12 \) and the semi-minor axis (along the y-axis) is \( 4 \).

3. **Graph the equations:**
- Draw the circle with radius \( 5 \) and the ellipse with semi-major axis \( 12 \) and semi-minor axis \( 4 \). The circle will be completely inside the ellipse since its radius is smaller than the semi-major axis of the ellipse.

4. **Find the points of intersection:**
- To find the points where the circle and the ellipse intersect, we can substitute \( y^2 \) from the circle's equation into the ellipse's equation.
- From the circle's equation, we have:
\[ y^2 = 25 - x^2 \]
- Substitute \( y^2 \) into the ellipse's equation:
\[ x^2 + 9(25 - x^2) = 144 \]
Simplifying this gives:
\[ x^2 + 225 - 9x^2 = 144 \]
\[ -8x^2 + 225 = 144 \]
\[ -8x^2 = 144 - 225 \]
\[ -8x^2 = -81 \]
\[ x^2 = \frac{81}{8} \]
\[ x = \pm \sqrt{\frac{81}{8}} = \pm \frac{9}{\sqrt{8}} = \pm \frac{9\sqrt{2}}{4} \]

5. **Find corresponding \( y \) values:**
- Substitute \( x^2 = \frac{81}{8} \) back into the circle's equation to find \( y^2 \):
\[ y^2 = 25 - \frac{81}{8} = \frac{200}{8} - \frac{81}{8} = \frac{119}{8} \]
\[ y = \pm \sqrt{\frac{119}{8}} = \pm \frac{\sqrt{119}}{2\sqrt{2}} = \pm \frac{\sqrt{238}}{4} \]

6. **Count the points of intersection:**
- Since \( x \) can take on two values (\( +\frac{9\sqrt{2}}{4} \) and \( -\frac{9\sqrt{2}}{4} \)) and \( y \) can also take on two values (\( +\frac{\sqrt{238}}{4} \) and \( -\frac{\sqrt{238}}{4} \)), we can find a total of \( 2 \times 2 = 4 \) points of intersection.

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