College Algebra (11th Edition)

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

Chapter 2 - Section 2.5 - Equations of Lines and Linear Models - Summary Exercises on Graphs, Circles, Functions, and Equations - Page 231: 21

Answer

does not have a circle as its graph

Work Step by Step

$\bf{\text{Solution Outline:}}$ To evaluate whether the given equation, $ x^2+y^2-2x+10=0 ,$ has a circle as its graph, change the form of the equation to the Center-Radius Form. Then evaluate the value of $r^2.$ If the value of $r^2$ is positive, then the graph of the equation is a circle. If $r^2$ is equal to zero, then the equation describes a point. If $r^2$ is negative, the graph is nonexistent. $\bf{\text{Solution Details:}}$ Grouping the $x$-variables and the $y$-variable and transposing the constant, the given equation is equivalent to \begin{array}{l}\require{cancel} (x^2-2x)+y^2=-10 .\end{array} Since the coefficient of $x$ is $ -2 ,$ add $\left(\dfrac{ -2 }{2}\right)^2= 1 $ to both sides of the equation to complete the square for $x$. That is, \begin{array}{l}\require{cancel} (x^2-2x+1)+y^2=-10+1 \\\\ (x-1)^2+y^2=-9 .\end{array} In the form $(x-h)^2+(y-k)^2=r^2,$ the equation above is equivalent to \begin{array}{l}\require{cancel} (x-(1))^2+(y-0)^2=-9 .\end{array} Since $r^2=-9,$ is less than zero, then the equation $\text{ does not have a circle as its graph }.$
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