College Algebra (11th Edition)

Published by Pearson
ISBN 10: 0321671791
ISBN 13: 978-0-32167-179-0

Chapter 2 - Section 2.2 - Circles - 2.2 Exercises: 26

Answer

graph of the equation is nonexistent

Work Step by Step

$\bf{\text{Solution Outline:}}$ To determine whether the given equation, $ x^2+y^2+4x-8y+32=0 ,$ is the graph of a circle, complete the square for both $x$ and $y$ variables and then analyze the value of the radius. If the value of the radius is positive, then the equation has a graph of a circle. If the radius is $0,$ then the equation describes a point. If the radius is negative, the graph is nonexistent. $\bf{\text{Solution Details:}}$ Grouping the $x$ variables and the $y$ variables and also transposing the constant, the given equation is equivalent to \begin{array}{l}\require{cancel} (x^2+4x)+(y^2-8y)=-32 .\end{array} With the coefficient of $x$ equal to $ 4 ,$ add $\left(\dfrac{ 4 }{2}\right)^2=4 $ to both sides of the equation above to complete the square for $x$. That is, \begin{array}{l}\require{cancel} (x^2+4x+4)+(y^2-8y)=-32+4 \\\\ (x+2)^2+(y^2-8y)=-28 .\end{array} With the coefficient of $y$ equal to $ -8 ,$ add $\left(\dfrac{ -8 }{2}\right)^2=16 $ to both sides of the equation above to complete the square for $y$. That is, \begin{array}{l}\require{cancel} (x+2)^2+(y^2-8y+16)=-28+16 \\\\ (x+2)^2+(y-4)^2=-12 .\end{array} Since $r^2=-12,$ is less than $0,$ the $\text{ graph of the equation is nonexistent }.$
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