/*! 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 29 Sketch the circle. Identify its ... [FREE SOLUTION] | 91影视

91影视

Log out

Learning Materials

Features

Discover

Chapter 10: Problem 29

Sketch the circle. Identify its center and radius. $$x^{2}+8 x+y^{2}+2 y+8=0$$

Short Answer

Expert verified
The center of the circle is at (-4, -1) and the radius of the circle is 3.

Step by step solution

01

Complete the square for x and y components in the circle equation

Rearrange the terms into groups and complete the square on the x-terms and y-terms. \[x^{2}+8x+y^{2}+2y=-8\] becomes \[(x^{2}+8x)+(y^{2}+2y)=-8\] which further simplifies to \[(x+4)^2 - 16 + (y+1)^2 - 1 = -8\]
02

Simplify the equation

Simplify the equation to the standard form of a circle equation by moving the constants to the right-hand side. \[(x+4)^2 - 16 + (y+1)^2 - 1 = -8\] can be simplified to \[(x+4)^2 + (y+1)^2 = 9\].
03

Identify the circle's parameters

The standard form of the equation is \((x-h)^2 + (y-k)^2 = r^2\). By comparing this to our simplified equation of \[(x+4)^2 + (y+1)^2 = 9\], we can identify \(h = -4\), \(k = -1\), and \(r = 3\).

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 a crucial technique used in algebra to transform quadratic equations into a simpler form. For a circle equation, it helps in identifying the center and the radius. Here is how you can complete the square:
  • Take a quadratic expression like \(x^2 + 8x\) or \(y^2 + 2y\).
  • For each quadratic expression, half the coefficient of the linear term, square it, and then add and subtract this value within the expression.
For example:
  • For \(x^2 + 8x\), half of 8 is 4, and 4 squared is 16, so it becomes \((x+4)^2 - 16\).
  • For \(y^2 + 2y\), half of 2 is 1, and 1 squared is 1, so it becomes \((y+1)^2 - 1\).
These steps convert the original equation into a format that reveals the circle鈥檚 center and radius clearly.
Standard Form of a Circle
The standard form of a circle's equation is essential to easily identify its geometrical properties. This form is \[(x-h)^2 + (y-k)^2 = r^2\]where the circle鈥檚 center is \((h, k)\) and \(r\) is the radius. After completing the square, the equation appeared as:
  • \((x+4)^2 + (y+1)^2 = 9\)
This form is directly comparable with the standard circle equation to extract the properties of the circle readily.
Center of a Circle
The center of a circle is one of the key features that helps position it within the coordinate plane. Once you have your equation in the standard form, identifying the center \((h, k)\) becomes effortless.
  • From the equation \((x+4)^2 + (y+1)^2 = 9\), compare it to \((x-h)^2 + (y-k)^2 = r^2\).
  • The terms \(x+4\) and \(y+1\) reveal the center coordinates as \(h = -4\) and \(k = -1\).
Thus, the circle's center is at \((-4, -1)\), guiding you both in graphical plotting and analysis.
Radius of a Circle
The radius of a circle is another fundamental characteristic, determining its size and scale within the coordinate space. Once the equation is in standard form, extracting the radius is straightforward.
  • The equation \((x+4)^2 + (y+1)^2 = 9\) shows the right side as \(9\).
  • Since the standard form is \((x-h)^2 + (y-k)^2 = r^2\), you find \(r^2 = 9\).
  • Solving for \(r\), you find that \(r = 3\).
Therefore, the radius of the circle is 3, which indicates its span from the center to any point on the circle's edge.

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

On November 27, \(1963,\) the United States launched a satellite named Explorer \(18 .\) Its low and high points above the surface of Earth were about 119 miles and 122,800 miles, respectively (see figure). The center of Earth is at one focus of the orbit. (a) Find the polar equation of the orbit (assume the radius of Earth is 4000 miles). (b) Find the distance between the surface of Earth and the satellite when \(\theta=60^{\circ}\). (c) Find the distance between the surface of Earth and the satellite when \(\theta=30^{\circ}\).

Convert the polar equation to rectangular form. $$\theta=\pi / 2$$

Convert the polar equation to rectangular form. $$r=\frac{1}{1-\cos \theta}$$

Convert the rectangular equation to polar form. Assume \(a<0\) $$x^{2}+y^{2}=16$$

Use a graphing utility to approximate any relative minimum or maximum values of the function. $$f(x)=x^{5}-3 x-1$$
See all solutions

Recommended explanations on Math Textbooks

Logic and Functions

Read Explanation

Decision Maths

Read Explanation

Mechanics Maths

Read Explanation

Statistics

Read Explanation

Applied Mathematics

Read Explanation

Discrete Mathematics

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.