Chapter 3: Problem 52
Exer. 47-56: Find the center and radius of the circle with the given equation. $$ 9 x^{2}+9 y^{2}+12 x-6 y+4=0 $$
Short Answer
Expert verified
The center is \((-\frac{2}{3}, \frac{1}{3})\) and the radius is 1.
Step by step solution
01
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: \(x^2 + y^2 + \frac{4}{3}x - \frac{2}{3}y + \frac{4}{9} = 0\).
02
Rearrange the terms
Reorganize the equation by grouping the \(x\) terms and \(y\) terms together: \((x^2 + \frac{4}{3}x) + (y^2 - \frac{2}{3}y) = -\frac{4}{9}\).
03
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: \((x^2 + \frac{4}{3}x + \frac{4}{9}) - \frac{4}{9}\).
04
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: \((y^2 - \frac{2}{3}y + \frac{1}{9}) - \frac{1}{9}\).
05
Rewrite the completed squares
Rewrite the equation using these completed squares: \((x + \frac{2}{3})^2 - \frac{4}{9} + (y - \frac{1}{3})^2 - \frac{1}{9} = -\frac{4}{9}\).
06
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: \((x + \frac{2}{3})^2 + (y - \frac{1}{3})^2 = 1\).
07
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 \(\sqrt{1} = 1\).
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Key Concepts
These are the key concepts you need to understand to accurately answer the question.
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.
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.
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.