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Chapter 0: Problem 20

Describe the set of points \((x, y)\) such that \(x^{2}+y^{2}=0\).

Short Answer

Expert verified
The set of points \((x, y)\) that satisfy the equation \(x^2+y^2=0\) is just a single point, which is \((0, 0)\). This result indicates that the circle has a radius of 0, basically being reduced to a single point at the center.

Step by step solution

01

1. Understand the equation

We are given the equation \(x^2+y^2=0\). This is the equation of a circle with center point at the origin (0,0) and a variable radius. In this particular equation, the sum of the square of x-coordinate and the square of y-coordinate is equal to 0.
02

2. Identify possible solutions

To find the points that satisfy this equation, we can set up the equation as follows: \(x^2+y^2=0\) Remember that for real numbers, a square of a number can never be negative, thus \(x^2 \geq 0\) and \(y^2 \geq 0\) In order to satisfy the equation \(x^2+y^2=0\), the only possibility is when both \(x^2=0\) and \(y^2=0\),
03

3. Find the point

Now, we can find the value of x and y that make \(x^2\) and \(y^2\) equal 0: If \(x^2=0\), then \(x=0\) If \(y^2=0\), then \(y=0\) So, the point that satisfies the given equation is \((x, y) = (0, 0)\).
04

4. Conclusion

The set of points \((x, y)\) that satisfy the equation \(x^2+y^2=0\) is just a single point, which is \((0, 0)\). This result indicates that the circle has a radius of 0, basically being reduced to a single point at the center.

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Most popular questions from this chapter

\(\left(x^{2}+1\right)^{5}(x+3)^{4}+\left(x^{2}+1\right)^{6}(x+3)^{3}\)

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