Problem 58 - Algebra and Trigonometry Real Mathematics, Real People

Question

Solve for \(y\) and use a graphing utility to graph each of the resulting equations in the same viewing window. (Adjust the viewing window so that the circle appears circular.) \((x-3)^{2}+(y-1)^{2}=25\)

Step-by-Step Solution

Answer

The isolations for \(y\) from the given circle equation are \(y = 1 + \sqrt{25 - (x-3)^{2}}\) and \(y = 1 - \sqrt{25 - (x-3)^{2}}\). Plotting these equations on a graph, adjusting the viewing window as required, will represent a circle centered at (3,1) with a radius of 5 units.

Step 1: Breakdown the given equation

Understand that the given equation \((x-3)^{2}+(y-1)^{2}=25\) is in the form \((x-a)^{2}+(y-b)^{2}=r^{2}\) where \((a,b)\) is the center of the circle and \(r\) is the radius. Here, the center is (3,1) and the radius \(r= \sqrt{25} = 5\). So this is a circle centered at (3,1) with a radius of 5 units.

Step 2: Solve for y

To solve for \(y\), subtract \((x-3)^{2}\) from both sides of the equation to isolate \(y\). The equation becomes:\(y-1 = \pm \sqrt{25-(x-3)^{2}}\). To isolate \(y\), add 1 to both sides which results in two equations for \(y\), namely \(y = 1 + \sqrt{25-(x-3)^{2}}\) and \(y = 1 - \sqrt{25-(x-3)^{2}}\).

Step 3: Graph the isolations

Be ready to graph the two isolated \(y\) equations using a graphing utility. Make sure to adjust the window so that the graph appears as a circle. The two equations represent the upper and lower halves of the circle respectively. The center as stated earlier is at (3,1), and the radius is 5 units.

Key Concepts

Solving for y

When working with circles in algebra, solving for y can be a bit tricky, because you often end up with a square root due to the nature of the circle equation. Taking our example, (x-3)^2 + (y-1)^2 = 25, the goal is to express y as a function of x. To achieve this, we rearranged the circle equation to isolate the y-term, leading us to two possible solutions because of the square root:
y = 1 + \[25-(x-3)^2\]^{0.5} and y = 1 - \[25-(x-3)^2\]^{0.5}.
This step is vital because it allows for the graphing of the equation in parts as the function is not one-to-one; it doesn't pass the vertical line test for functions. Thus, you graph the two separate parts—typically the upper and lower semicircles—based on the positive and negative roots.

Circle Equation Standard Form

The standard form of a circle's equation makes it convenient to identify the center and radius of the circle. This standard form is
(x-a)^2 + (y-b)^2 = r^2,
where

• (a, b) represents the center of the circle
• r is the radius of the circle

In the given exercise, we have
(x-3)^2 + (y-1)^2 = 25,
which is already in standard form. By comparing it to the standard form, we can immediately tell that the center is at (3,1) and, by taking the square root of 25, that the radius of the circle is 5 units. Understanding the standard form is key to quickly and effectively breaking down a circle's equation.

Using Graphing Calculators

Graphing calculators are invaluable tools for visualizing equations, especially when those equations are not straightforward lines. After solving for y, graphing the two parts of the circle is essential for a complete visual representation. It's important to adjust the viewing window to ensure the circle appears as it should—actual circles can look distorted if the window is not set correctly. For example, with a circle that has a radius of 5 units, you might set the window to have a range of at least -10 to 10 on both axes to view the entire structure. With the solved equations for y, use the graphing calculator to plot both the upper and lower halves separately, ensuring the circle is visualized perfectly.

Radius and Center of A Circle

The radius and center of a circle are foundational concepts in geometry and are easily identified from the equation in standard form. The center of the circle is simply the point (a, b) in the standard form equation. Understanding the center allows students to visualize where on the coordinate plane the circle will be positioned.
The radius is the distance from the center of the circle to any point on its circumference and is represented by r in our standard form equation. For our circle with equation
(x-3)^2 + (y-1)^2 = 25,
the radius, r, is 5, as it's the square root of 25. Together, the center and radius describe the size and location of the circle, which is crucial for graphing and understanding the shape's properties algebraically.