Explanations 
Textbooks 
Flashcards 
91影视 AI 
Notes 
Study Plans 
Study Sets 
Find a degree 
Magazine Chapter 8: Problem 43
\(x^{2}+y^{2}+4 x+4 y-8=0\)
Short Answer
Expert verified
Circle with center (-2, -2) and radius 4.
Step by step solution
01
Rewrite the Equation
Reposition terms to see the equation as a sum of squares. The given equation is \(x^2 + y^2 + 4x + 4y - 8 = 0\). We will rearrange it to: \(x^2 + 4x + y^2 + 4y = 8\).
02
Complete the Square for x
To complete the square for \(x\), focus on the terms \(x^2 + 4x\). Add \(4\) (from \((\frac{4}{2})^2 = 4\)) to both sides of the equation: \((x + 2)^2 \). So, \( (x + 2)^2 - 4 + y^2 + 4y = 8 \).
03
Complete the Square for y
Now complete the square for \(y\) within the equation. Focus on \(y^2 + 4y\). Add \(4\) (since \((\frac{4}{2})^2 = 4\)) to transform it to \((y + 2)^2 \). Thus, \((x + 2)^2 - 4 + (y + 2)^2 - 4 = 8 \).
04
Simplify the Equation
Now, simplify the equation: \((x + 2)^2 + (y + 2)^2 - 8 = 8\). Move \(-8\) from the left side to the right side to get \((x + 2)^2 + (y + 2)^2 = 16\).
05
Identify the Circle's Center and Radius
The equation \((x + 2)^2 + (y + 2)^2 = 16\) represents a circle in standard form \((x - h)^2 + (y - k)^2 = r^2\). From this, we identify that the center: \((-2, -2)\) and the radius: \(4\) (since \(r^2 = 16\), \(r = \sqrt{16} = 4\)).
Unlock Step-by-Step Solutions & Ace Your Exams!
-
Full Textbook Solutions
Get detailed explanations and key concepts
-
Unlimited Al creation
Al flashcards, explanations, exams and more...
-
Ads-free access
To over 500 millions flashcards
-
Money-back guarantee
We refund you if you fail your exam.
Over 30 million students worldwide already upgrade their learning with 91影视!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Circle Equation
In geometry, a circle is a two-dimensional shape consisting of all points that are equidistant from a fixed center point. The equation of a circle on a coordinate plane can take various forms, but one of the most common is the standard form, which makes it easier to identify key features like the center and radius.
- The general equation for a circle is \(x^2 + y^2 + Dx + Ey + F = 0\).
- Transforming this into the standard form \( (x - h)^2 + (y - k)^2 = r^2\), enables us to easily spot the center \( (h, k)\) and the radius \( r\).
- Our specific example transforms to \( (x + 2)^2 + (y + 2)^2 = 16\).
Standard Form
The standard form of a circle equation is \( (x - h)^2 + (y - k)^2 = r^2\). This format is very helpful because it allows you to quickly identify key circle components. To transform a general equation into standard form, completing the square is often necessary:
- Binomials like \(x^2 + 4x\) and \(y^2 + 4y\) are converted into perfect squares: \( (x + 2)^2\) and \( (y + 2)^2\).
- The number subtracted from each squared term represents the square completed during the transformation process.
- The constant term on the right, like 16, tells you that \( r^2 = 16\) which means \( r = 4\).
Algebraic Manipulation
Algebraic manipulation is a crucial process in transforming equations to reveal more informative insights. For circle equations, these techniques help express the original equation in a form that highlights important properties. Let's break down some algebraic tips:
- First, isolate terms involving \( x \) and \( y \) and focus on completing the square, which requires adding \( \left( \frac{b}{2} \right)^2 \) to both sides where \( b \) is the linear coefficient.
- Our exercise used \( (\frac{4}{2})^2 = 4 \) for both \( x \) and \( y \) components.
- Rearranging and balancing the equation involves moving terms around so that \( (x + 2)^2 + (y + 2)^2 = 16\) was reached efficiently.
