Problem 77 - College Algebra and Calculus: An Applied Approach

Question

Sketch the graph of the equation. Identify any intercepts and test for symmetry. \(x^{2}+y^{2}=4\)

Step-by-Step Solution

Answer

The graph of the equation \(x^{2}+y^{2}=4\) is a circle centered at the origin (0,0) with a radius of 2. The intercepts are at (±2, 0) and (0, ±2). The graph is symmetric with respect to the x-axis, y-axis, and origin.

Step 1: Identify the form of the equation

The equation \(x^{2}+y^{2}=4\) is in the standard form for the equation of a circle centered at the origin, which is \(x^{2}+y^{2}=r^{2}\). It signifies a circle with a radius, \(r\), of 2 (since \(\sqrt{4} = 2\)) and center at the origin (0,0).

Step 2: Determine the intercepts

To determine the intercepts, set each variable equal to zero and solve for the other variable. For the x-intercepts, substitute \(y=0\) into the equation and solve for \(x\), the outcome is \(x = ± \sqrt{4} = ±2\). The x-intercepts are therefore (2,0) and (-2,0). For the y-intercepts, substitute x=0 into the equation and solve for y, which gives \(y = ± \sqrt{4} = ±2\). The y-intercepts are therefore (0,2) and (0,-2).

Step 3: Test for symmetry

The graph of a circle is symmetrical about both the x-axis and the y-axis. Indeed, replacing \(x\) with \(-x\) and \(y\) with \(-y\) in the equation results in the same equation, confirming that the graph is symmetric about both the x-axis and y-axis.

Step 4: Sketch the graph

The graph of the equation \(x^{2}+y^{2}=4\) is a circle with the center at the origin (0,0) and a radius of 2. Simply plot the center, then from the center plot the four intercepts (±2,0) and (0,±2). You can then complete the circle by sketching it around these points.

Key Concepts

Equation of a Circle

A circle's equation in mathematics can be wonderfully simple. It is typically written in its standard form as \(x^{2}+y^{2}=r^{2}\), where \(r\) is the radius of the circle and the center is at the origin (0,0). This form tells us two critical components of any circle: its center point and its radius. In the given equation \(x^{2}+y^{2}=4\), we immediately recognize the circle's center is at the origin, while the radius is the square root of 4, which is 2. This means the circle extends 2 units away from the center in all directions. To sum it up:

• The circle's center: (0,0)
• The radius: 2

Knowing these, you can easily plot the foundational points of the circle on a graph.

Intercepts

Intercepts are where the circle crosses the x-axis and y-axis. Finding these points helps in sketching accurate graphs because they inform where the shape touches the axes. To find the x-intercepts:

• Set \(y = 0\) in the equation \(x^{2}+y^{2}=4\), which simplifies to \(x^{2} = 4\).
• Solve for \(x\) to get \(x = \pm 2\), so the x-intercepts are (2,0) and (-2,0).

To find the y-intercepts:

• Set \(x = 0\) in the equation \(x^{2}+y^{2}=4\), which simplifies to \(y^{2} = 4\).
• Solve for \(y\), which yields \(y = \pm 2\), so the y-intercepts are (0,2) and (0,-2).

Identifying these intercepts allows you to anchor the circle on a graph, making it a crucial step in visualizing the circle.

Symmetry Testing

Symmetry makes graphing much easier by simplifying the computations involved in sketching out circles. A circle has inherent symmetry, and in the equation \(x^{2}+y^{2}=4\), this symmetry is evident. To test for symmetry, replace \(x\) with \(-x\) and \(y\) with \(-y\) in the equation.- Replacing \(x\) with \(-x\) gives: \((-x)^{2}+y^{2} = 4\), which simplifies back to \(x^{2} + y^{2} = 4\).- Replacing \(y\) with \(-y\) gives: \(x^{2}+(-y)^{2} = 4\), simplifying again to \(x^{2} + y^{2} = 4\).Both transformations demonstrate that the original equation remains unchanged, which confirms symmetry in respect to both the x-axis and the y-axis. This means the circle not only looks the same above and below the x-axis but also to the left and right of the y-axis. Understanding this concept helps ensure your graph is accurate and that the circle is consistent across all quadrants.