Problem 52 - Algebra and Trigonometry with Analytic Geometry

Question

Exer. 47-56: Find the center and radius of the circle with the given equation. $$ 9 x^{2}+9 y^{2}+12 x-6 y+4=0 $$

Step-by-Step Solution

Answer

The center is \((-\frac{2}{3}, \frac{1}{3})\) and the radius is 1.

Step 1: Simplify the equation

Begin by dividing the entire equation by 9 to make the coefficients of \(x^2\) and \(y^2\) equal to 1, thus: \(x^2 + y^2 + \frac{4}{3}x - \frac{2}{3}y + \frac{4}{9} = 0\).

Step 2: Rearrange the terms

Reorganize the equation by grouping the \(x\) terms and \(y\) terms together: \((x^2 + \frac{4}{3}x) + (y^2 - \frac{2}{3}y) = -\frac{4}{9}\).

Step 3: Complete the square for x

To complete the square for the \(x\) terms, take half of \(\frac{4}{3}\), square it, and add it inside the bracket. So, half of \(\frac{4}{3}\) is \(\frac{2}{3}\) and \(\left(\frac{2}{3}\right)^2 = \frac{4}{9}\). Add and subtract \(\frac{4}{9}\) inside the \(x\) group: \((x^2 + \frac{4}{3}x + \frac{4}{9}) - \frac{4}{9}\).

Step 4: Complete the square for y

Similarly, complete the square for the \(y\) terms. Take half of \(-\frac{2}{3}\), square it, and add it inside the bracket. Half of \(-\frac{2}{3}\) is \(-\frac{1}{3}\), and \(\left(-\frac{1}{3}\right)^2 = \frac{1}{9}\). Add and subtract \(\frac{1}{9}\) in the \(y\) group: \((y^2 - \frac{2}{3}y + \frac{1}{9}) - \frac{1}{9}\).

Step 5: Rewrite the completed squares

Rewrite the equation using these completed squares: \((x + \frac{2}{3})^2 - \frac{4}{9} + (y - \frac{1}{3})^2 - \frac{1}{9} = -\frac{4}{9}\).

Step 6: Simplify and find the circle equation

Combine all the constants on one side: \((x + \frac{2}{3})^2 + (y - \frac{1}{3})^2 = \frac{9}{9}\). This simplifies to: \((x + \frac{2}{3})^2 + (y - \frac{1}{3})^2 = 1\).

Step 7: Identify the center and radius

From the equation \((x + \frac{2}{3})^2 + (y - \frac{1}{3})^2 = 1\), the center of the circle is \((-\frac{2}{3}, \frac{1}{3})\) and the radius is \(\sqrt{1} = 1\).

Key Concepts

Completing the Square

Completing the square is a method used in algebra to shift quadratic expressions into a form where they can be analyzed more simply. This is especially useful when working with circles, as it allows us to express circle equations in a more recognizable form:

• Start by organizing terms: Identify and group the individual \(x\) and \(y\) terms.
• Next, transform each grouped term into a perfect square trinomial by adding and subtracting the same value.

Let's consider two main steps:1. **Half the Coefficient, and Square It**: Take half of the linear term's coefficient and square it. For example, if you have \(x^2 + \frac{4}{3}x\), half of \(\frac{4}{3}\) is \(\frac{2}{3}\), and squared is \(\frac{4}{9}\).2. **Add and Subtract Inside**: Add this square inside the group, ensuring the equation's value doesn't change by also subtracting it immediately after.This process transforms expressions into perfect square form, which makes it simple to identify the circle's properties.

Center and Radius of a Circle

A circle's equation in standard form reveals valuable information about the circle's geometry. When the equation is finally expressed as \((x - h)^2 + (y - k)^2 = r^2\), it becomes straightforward to discover the circle's center and radius:

• **Center**: Given by the coordinates \((h, k)\), which are derived from the transformed terms \((x + \frac{2}{3})^2\) and \((y - \frac{1}{3})^2\). Remember, the expression \((x + a)^2\) implies a center coordinate of \(-a\).
• **Radius**: The radius is \(r\), obtained from taking the square root of the constant on the equation's right side. In our example, with \((x + \frac{2}{3})^2 + (y - \frac{1}{3})^2 = 1\), the radius is simply \(1\), because \(\sqrt{1} = 1\).

Coordinate Geometry

Coordinate geometry, also known as analytic geometry, merges algebra and geometry using a coordinate system. This is crucial when working with circle equations:

• **Cartesian Plane**: Circles are visualized on a plane with axes (x and y), making it easy to plot shapes using their equations.
• **Coordinates**: Designating exact positions of points (\(x, y\)) helps us precisely locate the circle's center and accurately depict its radius's reach.

This mathematical field simplifies complex geometric figures into understandable visual representations through equations, like placing a circle accurately on its plane.

Standard Form of a Circle

The standard form of a circle equation is essential for easily interpreting circles. It is written as \((x - h)^2 + (y - k)^2 = r^2\):

• **Easy Reading**: The form directly gives the center \((h,k)\) and radius \(r\).
• **Efficiency**: By reorganizing the equation to this form, complex algebra is reduced, making geometric conclusions straightforward.
• **Conversion**: Transformation from general form to standard helps in quickly identifying circle properties from any equation. The general form of circle equations (e.g., \(x^2 + y^2 + Dx + Ey + F = 0\)) sometimes requires completing the square to convert.

Understanding the standard form is critical for analyzing and using circle equations confidently across various applications in mathematics and science.