Problem 84 - College Algebra Essentials

Question

use a graphing utility to graph each circle whose equation is given. $$ (y+1)^{2}=36-(x-3)^{2} $$

Step-by-Step Solution

Answer

The circle's center is at the point (3, -1) and its radius is 6 units. So, the graph of the circle will be a perfect circle centered at (3, -1) with a radius of 6 units.

Step 1: Identify the Standard Form of the Circle

In this exercise, the standard form of the circle's equation is \((x-h)^{2}+(y-k)^{2}=r^{2}\). Comparing this with the given equation \((y+1)^{2}=36-(x-3)^{2}\), rewrite the equation as \((x-3)^{2}+(y+1)^{2}=36\). Here, \((h, k)\) are the coordinates of the circle's center and \(r\) is the radius. So we understand that the circle's center is at (3, -1) and the radius is \(\sqrt{36}\) = 6 units.

Step 2: Plot the Circle's Center

Use a graphing utility (such as a graphing calculator or computer software) to mark the center of the circle at point (3, -1).

Step 3: Draw the Circle

Using the graphing utility, draw a circle with the center at point (3, -1) and the radius as 6 units.

Key Concepts

Standard Form of a Circle

Understanding the standard form of a circle is crucial for graphing and comprehending circle equations. The standard form is written as \((x-h)^{2}+(y-k)^{2}=r^{2}\). Here, \((h, k)\) represents the center of the circle, while \(r\) is the radius.

In the equation \((x-h)^{2}+(y-k)^{2}=r^{2}\):

• \((x-h)^{2}\) and \((y-k)^{2}\) show how each point \((x, y)\) on the circle relates to the center.
• Subtracting \(h\) and \(k\) from \(x\) and \(y\) shifts the focus to the center.
• Squaring the terms ensures all points at a distance \(r\) from the center are included.

For our exercise, the initial equation needs rearranging to match the standard form: \((x-3)^{2} + (y+1)^{2} = 36\).

This transformation helps us to easily plot or analyze circles by identifying the center and the radius right from the equation.

Graphing Utility

A graphing utility simplifies the visualization of the circle by utilizing digital tools. These utilities can be graphing calculators or software programs like Desmos or GeoGebra.

Using a graphing utility involves a few straightforward steps:

• Input the standard form equation of the circle into the utility.
• Graph the equation by specifying the center and radius.
• Modify graph settings to fit the circle within the viewable region.

By following these steps using the equation \((x-3)^{2} + (y+1)^{2} = 36\), the circle can be drawn perfectly with a center at \((3, -1)\) and a radius of 6 units. Graphing utilities also allow for easy adjustments and visual confirmation of the circle's properties.

Circle's Center and Radius

The center and radius are fundamental components of a circle's geometry. Identifying these from the equation allows for easy graphing and deeper understanding.

In the standard circle equation \((x-h)^{2} + (y-k)^{2} = r^{2}\):

• The center \((h, k)\) is determined by the values inside the parentheses. Here, \((h, k)\) is \((3, -1)\).
• The radius \(r\) is the square root of the number on the right side of the equation. For our circle, \(r = \sqrt{36} = 6\).

Knowing the center as \((3, -1)\) and radius as 6, one can accurately plot the circle. The center is the anchor point, and the radius represents the distance from this center to any point on the circle's edge. By understanding these concepts, solving and visualizing circle equations becomes intuitive and manageable.