Problem 14 - A Graphical Approach to College Algebra

Question

Find the center-radius form for each circle satisfying the given conditions. $$\text { Center }\left(-\frac{1}{2},-\frac{1}{4}\right) ; \text { radius } \frac{12}{5}$$

Step-by-Step Solution

Answer

\((x + \frac{1}{2})^2 + (y + \frac{1}{4})^2 = \frac{144}{25}\)

Step 1: Understand the Circle Equation

The standard (or center-radius) form of a circle's equation is given by \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) is the center of the circle and \(r\) is the radius.

Step 2: Identify Given Values

We're given the center of the circle \((-\frac{1}{2}, -\frac{1}{4})\) and the radius \(\frac{12}{5}\). So, \(h = -\frac{1}{2}\), \(k = -\frac{1}{4}\), and \(r = \frac{12}{5}\).

Step 3: Substitute into the Circle Equation

Substitute the given values into the standard circle equation:\[(x - (-\frac{1}{2}))^2 + (y - (-\frac{1}{4}))^2 = \left(\frac{12}{5}\right)^2\]

Step 4: Simplify the Equation

Simplify the equation from the previous step:1. Change the double negatives to positives: \((x + \frac{1}{2})^2 + (y + \frac{1}{4})^2 = \left(\frac{12}{5}\right)^2\).2. Square the radius: \[\left(\frac{12}{5}\right)^2 = \frac{144}{25}\].So the equation becomes \((x + \frac{1}{2})^2 + (y + \frac{1}{4})^2 = \frac{144}{25}\).

Key Concepts

Center-Radius Form of a Circle

The center-radius form is a simplified way to express the equation of a circle. The formula for this form is \((x - h)^2 + (y - k)^2 = r^2\), where \((h, k)\) represents the center point of the circle, and \(r\) is the length of the radius. This form is quite helpful because it provides direct information about the circle's key properties - its center and radius - making it easier to visualize and work with.

• \(h\) and \(k\) are the x and y coordinates of the center.
• \(r\) is the radius.
• Equation directly relates to the circle’s distance from the center to any point on the circle's edge.

Understanding this structure is crucial as it helps form the foundation for working with circles in algebra. By knowing the center and radius, you can quickly sketch or manipulate the circle in further problems.

Standard Form of a Circle Equation

When dealing with circles, the standard form, especially the center-radius form, doesn't just help in identifying the circle's properties, but also aids in performing calculations. For example, with the circle having the center at \((-\frac{1}{2}, -\frac{1}{4})\) and radius \(\frac{12}{5}\), substituting these values into the center-radius equation gives you the circle's specific form.
In this context:

• The equation starts as \((x - (-\frac{1}{2}))^2 + (y - (-\frac{1}{4}))^2 = \left(\frac{12}{5}\right)^2\).
• This can then be simplified to \((x + \frac{1}{2})^2 + (y + \frac{1}{4})^2 = \frac{144}{25}\).

Grasping the standard form will make identifying and solving circle-related problems more straightforward, whether you're completing square forms or converting between circle and other geometric forms.

Algebraic Substitution

Algebraic substitution is a powerful tool when working with equations, such as those involving circles. Substitution allows you to replace variables with given values to simplify equations and solve them. Specifically, when you're given the center \((-\frac{1}{2}, -\frac{1}{4})\) and radius \(\frac{12}{5}\), these values are directly substituted into the equation of the circle.
This involves:

• Replacing \(h\) with \(-\frac{1}{2}\) and \(k\) with \(-\frac{1}{4}\) in \(((x - h)^2 + (y - k)^2 = r^2)\).
• Substituting \(r\) with \(\frac{12}{5}\), hence \(r^2 = \left(\frac{12}{5}\right)^2\).

The skill of substitution not only simplifies the process but also helps ensure accurate solutions in algebra and geometry.

Simplification of Equations

Simplifying equations is an essential practice in many areas of math, helping to make complex problems more manageable. In the context of working with the equation of a circle, simplification involves generating a cleaner, more interpretable form of the expression and thus a more clear description of the circle.
During simplification:

• Negative signs, like in \((x - (-\frac{1}{2}))^2\), are turned to positives as \((x + \frac{1}{2})^2\) to eliminate double negatives.
• The radius \(\frac{12}{5}\) is squared to produce \(\frac{144}{25}\), simplifying calculations and providing a neat result.

By simplifying the equation, \((x + \frac{1}{2})^2 + (y + \frac{1}{4})^2 = \frac{144}{25}\), you arrive at a representation that is not only easier to work with but also readily usable in further analysis or graphic tasks.