Chapter 2: Problem 70
65–72 ? Show that the equation represents a circle, and find the center and radius of the circle. $$ x^{2}+y^{2}+2 x+y+1=0 $$
Short Answer
Expert verified
The circle's center is \((-1, -\frac{1}{2})\), and its radius is \(\frac{\sqrt{3}}{2}\).
Step by step solution
01
Recognize the Standard Form of a Circle
The standard form equation of a circle is \((x-h)^2 + (y-k)^2 = r^2\), where \((h, k)\) is the center and \(r\) is the radius of the circle.
02
Complete the Square for x-terms
Given the equation \(x^2 + y^2 + 2x + y + 1 = 0\), start by grouping the \(x\)-terms: \(x^2 + 2x\). To complete the square for \(x^2 + 2x\), add and subtract \((\frac{2}{2})^2 = 1\):\[(x^2 + 2x + 1) - 1 = (x+1)^2 - 1.\]
03
Complete the Square for y-terms
Now, group the y-terms: \(y^2 + y\). To complete the square, add and subtract \((\frac{1}{2})^2 = \frac{1}{4}\):\[(y^2 + y + \frac{1}{4}) - \frac{1}{4} = (y+\frac{1}{2})^2 - \frac{1}{4}.\]
04
Rewrite the Equation with Completed Squares
Now rewrite the entire equation using completed squares: \((x+1)^2 - 1 + (y+\frac{1}{2})^2 - \frac{1}{4} + 1 = 0.\) Simplify by combining constants on the right side of the equation:\[(x+1)^2 + (y+\frac{1}{2})^2 = 1 - \frac{1}{4} - 1 = \frac{3}{4}.\]
05
Identify the Center and Radius
The equation \((x+1)^2 + (y+\frac{1}{2})^2 = \frac{3}{4}\) is now in standard form. Hence, the center of the circle is \((-1, -\frac{1}{2})\), and the radius is \(\sqrt{\frac{3}{4}} = \frac{\sqrt{3}}{2}.\)
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 91Ó°ÊÓ!
Key Concepts
These are the key concepts you need to understand to accurately answer the question.
Completing the Square
Completing the square is a key algebraic technique for rewriting a quadratic equation. It helps in identifying certain patterns and can turn a complex equation into an easier-to-solve form.
When we are given a quadratic equation in the form of
This simplification is crucial when working with conics, such as circles, because it reveals the hidden structure within the equation.
When we are given a quadratic equation in the form of
- \( x^2 + bx \) or
- \( y^2 + cy \)
- \(( \frac{b}{2} )^2\) for the x-terms, and
- \(( \frac{c}{2} )^2\) for the y-terms.
This simplification is crucial when working with conics, such as circles, because it reveals the hidden structure within the equation.
Standard Form of a Circle
The standard form of a circle's equation is a neat representation that makes it easy to locate the circle's center and radius. The general form is \((x-h)^2 + (y-k)^2 = r^2\). In this form:
Once in the standard form, it clearly shows the geometrical nature of the circle, making it a powerful tool in solving problems related to circles.
- \((h, k)\) represents the circle's center
- \(r\) is the radius
Once in the standard form, it clearly shows the geometrical nature of the circle, making it a powerful tool in solving problems related to circles.
Center and Radius of a Circle
The center and the radius are defining properties of a circle. In an equation of the form \((x-h)^2 + (y-k)^2 = r^2\), it becomes straightforward to extract the circle's fundamental details.
Whether you're plotting the circle on a graph or analyzing its properties, the center and radius provide vital information about its size and position within coordinate geometry.
- The center
- The radius
Whether you're plotting the circle on a graph or analyzing its properties, the center and radius provide vital information about its size and position within coordinate geometry.