/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 41 (a) identify the center and radi... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 11: Problem 41

(a) identify the center and radius and (b) graph. $$ x^{2}+y^{2}+2 x+6 y+9=0 $$

Short Answer

Expert verified
Center: \((-1, -3)\), Radius: \(1\).Graph with center at \((-1, -3)\) and radius 1.

Step by step solution

01

Rewrite the equation in standard form

The given equation is: \[x^2 + y^2 + 2x + 6y + 9 = 0\].First, we need to rewrite this equation in the standard form of a circle equation, \[(x - h)^2 + (y - k)^2 = r^2\], by completing the square for both the x and y terms.
02

Rearrange and complete the square for x

Group the x terms together: \[x^2 + 2x\]. To complete the square: \[x^2 + 2x + 1 = (x + 1)^2\]. We add and subtract 1 within the equation.
03

Rearrange and complete the square for y

Group the y terms together: \[y^2 + 6y\]. To complete the square: \[y^2 + 6y + 9 = (y + 3)^2\]. We add and subtract 9 within the equation.
04

Substitute back into the equation

Combine the completed squares and simplify:\[x^2 + 2x + 1 + y^2 + 6y + 9 = -9 + 1 + 9\].This simplifies to: \[(x + 1)^2 + (y + 3)^2 = 1\].
05

Identify center and radius

The equation \[(x + 1)^2 + (y + 3)^2 = 1\] is in the standard form \[(x - h)^2 + (y - k)^2 = r^2\], where \(h = -1\), \(k = -3\), and \(r = 1\).Thus, the center is \((-1, -3)\) and the radius is \(1\).
06

Graph the circle

Plot the center of the circle at \((-1, -3)\). From this point, use the radius \(1\) to mark points one unit away in all directions: left, right, up, and down. Draw the circle through these points.

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.

Completing the Square
Completing the square is an essential algebra technique to transform a quadratic equation into a perfect square trinomial. This method is particularly useful in graphing circles.
Let's break down the process:
  • First, identify the quadratic terms and their coefficients. In the given problem, the x terms are \(x^2 + 2x\) and the y terms are \(y^2 + 6y\).
  • To complete the square for the x terms, add and subtract the square of half the coefficient of x. In our case, \(2 \times (1/2) = 1\). Thus, \(x^2 + 2x\) becomes \((x + 1)^2 - 1\).
  • Similarly, for the y terms, half the coefficient of y is \(6/2 = 3\), and its square is 9. Therefore, \(y^2 + 6y\) becomes \((y + 3)^2 - 9\).
By adding and then subtracting these perfect squares, we can rewrite the original equation in a simplified form. This is a crucial step to convert and simplify the equation of a circle.
Standard Form of a Circle Equation
The standard form of a circle's equation is \((x - h)^2 + (y - k)^2 = r^2\).
  • This format makes it easy to identify the circle's center and radius immediately.
  • The values \(h\) and \(k\) represent the coordinates of the center, while \(r\) is the radius.
  • The equation \(x^2 + y^2 + 2x + 6y + 9 = 0\) needs to be transformed into this standard form to graph the circle efficiently.
By completing the square for both x and y terms, and rearranging the given equation, we convert it to \((x + 1)^2 + (y + 3)^2 = 1\). This conversion simplifies the process of identifying the circle's fundamental properties and facilitates easier graphing.
Identifying Center and Radius
In the equation \((x + 1)^2 + (y + 3)^2 = 1\), the circle's center and radius are readily identified from the standard form:
  • The form \( (x - h)^2 + (y - k)^2 = r^2 \) tells us the center \((h, k)\) and the radius \(r\).
  • For \((x + 1)^2 + (y + 3)^2 = 1\), comparing with the standard form, we get \(h = -1\) and \(k = -3\). Therefore, the center is at \((-1, -3)\).
  • The expression \(r^2\) equals 1, so the radius \(r\) is the square root of 1, which is 1.
With the center at \(-1, -3\) and radius 1, we can now easily plot the circle by marking the center and drawing an equal distance all around.

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Graph each ellipse. $$ \frac{x^{2}}{16}+\frac{y^{2}}{36}=1 $$

(a) write the equation in standard form and (b) graph. $$ 4 x^{2}+25 y^{2}-24 x-64=0 $$

(a) write the equation in standard form and (b) graph. $$ 4 y^{2}-16 x^{2}-24 y+96 x-172=0 $$

Solve the problem using a system of equations. The difference of the squares of two numbers is 20. The difference of the square of the first number and twice the square of the second number is 4. Find the numbers.

Solve the system of equations by using graphing. $$ \left\\{\begin{array}{l} y=2 x+2 \\ y=-x^{2}+2 \end{array}\right. $$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Calculus

Read Explanation

Discrete Mathematics

Read Explanation

Statistics

Read Explanation

Theoretical and Mathematical Physics

Read Explanation

Probability and Statistics

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.