Question

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

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

Hint:

A circle and an ellipse can intersect at 0, 1, 2, 3, or 4 points.

Answer 1

The Correct Option is D

 To find the intersection of two sets \(A\) and \(B\), we must determine the points \((x, y)\) that satisfy both equations.

1. The first equation, \(x^2 + y^2 = 25\), represents a circle centered at the origin with a radius of 5.
2. The second equation, \(x^2 + 9y^2 = 144\), is an ellipse centered at the origin. To simplify, divide through by 144 to get it in standard form: \(\frac{x^2}{144} + \frac{9y^2}{144} = 1\), which simplifies to \(\frac{x^2}{144} + \frac{y^2}{16} = 1\).

Now we need to find the intersection points by solving these equations simultaneously:

1. Substitute \(x^2 = 25 - y^2\) from the circle into the ellipse's equation:
2. \(\frac{25 - y^2}{144} + \frac{y^2}{16} = 1\)
3. Simplify and solve for \(y^2\). To do this, first form a common denominator and solve for \(y^2\):
4. \(\frac{25 - y^2}{144} + \frac{9y^2}{144} = 1\) becomes \(\frac{25 + 8y^2}{144} = 1\)
5. This simplifies to: \(25 + 8y^2 = 144\)
6. Solve for \(y^2\):
7. \(8y^2 = 119 \, \Rightarrow \, y^2 = \frac{119}{8}\)

Substitute \(y^2\) back into the circle's equation to find \(x^2\):

1. \(x^2 = 25 - \frac{119}{8}\)
2. This results in \(x^2 = \frac{200}{8} - \frac{119}{8} = \frac{81}{8}\)

Now that we have \(x^2 = \frac{81}{8}\) and \(y^2 = \frac{119}{8}\), the possible solutions are the points \((\pm \frac{\sqrt{81}}{\sqrt{8}}, \pm \frac{\sqrt{119}}{\sqrt{8}})\).

Thus, the intersection results in four distinct points since each of \(x\) and \(y\) has two possible values (positive and negative). Therefore, \(A \cap B\) contains four points.

Hence, the correct answer is "four points."