Problem 19 - A Graphical Approach to College Algebra

Question

Find the center-radius form for each circle satisfying the given conditions. Describe the graph of the following equation. $$ (x-3)^{2}+(y-3)^{2}=0 $$

Step-by-Step Solution

Answer

The graph is a single point at (3, 3).

Step 1: Recognize the Circle Equation Format

The equation \((x-3)^2 + (y-3)^2 = 0\) is in the standard form of a circle equation \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center of the circle, and \(r\) is the radius.

Step 2: Identify the Center of the Circle

In this equation, \((h, k) = (3, 3)\), so the center of the circle is at \((3, 3)\).

Step 3: Determine the Radius

The equation \((x-3)^2 + (y-3)^2 = 0\) represents a radius squared of 0. Therefore, we take the square root of 0 to find that the radius \(r = 0\).

Step 4: Describe the Graph

A circle with a center at \((3, 3)\) and a radius of 0 is actually not a circle; it's a single point located at \((3, 3)\).

Key Concepts

Circle Equation

The circle equation is a fundamental concept in geometry that provides a formula to represent a circle on a coordinate plane. In general, the equation of a circle is given in the form

• \[(x-h)^2 + (y-k)^2 = r^2\]

where

• \((h, k)\) represents the center of the circle,
• and \(r\) represents the radius of the circle.

The equation uses a set of coordinates \((x, y)\) and calculates how they relate to the center point \((h, k)\) and the radius \(r\).
This formula is crucial for graphing a circle and understanding its geometric properties on a Cartesian plane.
Each point \((x, y)\) on the circle is equidistant from the center, making it symmetrical and round in shape.

Standard Form

The standard form of a circle's equation is expressed as \((x-h)^2+(y-k)^2=r^2\).
This form provides a straightforward way to read off the circle's properties, such as its center and radius.
When the equation is written in standard form:

• The numbers \(h\) and \(k\) can be read directly from the equation as the circle's center,
• and \(r^2\) gives the radius squared.

This format simplifies many calculations, allowing easy manipulation to solve for different properties of the circle.
In more complex scenarios, equations may require rearranging to fit this standard form, but once achieved, the visualization and interpretation of the circle's characteristics become much simpler.

Equation of a Circle

The equation of a circle is a vital tool in understanding circular shapes and their properties on a graph. A few key points about these equations include:

• The equation defines a boundary or perimeter where points satisfy \((x-h)^2 + (y-k)^2 = r^2\).
• When solved, it identifies all the points \((x, y)\) that lie on the circle.
• Graphically, a circle with this equation appears as a closed loop, except when \(r=0\) in which it reduces to a single point.

In practice, this equation helps visualize problems involving circles, whether determining intersections, tangents, or areas enclosures.
With this understanding, interpreting equations like

• \((x-3)^2 + (y-3)^2 = 0\)

becomes straightforward, quickly revealing

• the center,
• and noting its atypical graph as a single point at \((3,3)\), due to the radius being zero.

Center of a Circle

The center of a circle

• is a crucial point from which the circle is perfectly symmetrical.

In the circle equation, it is denoted by \((h, k)\) within the standard form

• \[(x-h)^2 + (y-k)^2 = r^2\].

Understanding the center's location is essential as it affects how the circle is positioned in the coordinate plane.
For instance, with the equation

• \((x-3)^2 + (y-3)^2 = 0\),

we directly interpret that the center is

• at the coordinates \((3,3)\).

Notably, when the radius is zero as in this case, the entire "circle" exists at just this location, collapsing it to a single point.
Thus, the center provides a reference point for drawing and understanding circular figures, even when atypical, like a zero-radius circle.