College Algebra 7th Edition

Published by Brooks Cole
ISBN 10: 1305115546
ISBN 13: 978-1-30511-554-5

Chapter 1, Equations and Graphs - Section 1.2 - Graphs of Equations in Two Variables; Circles - 1.2 Exercises - Page 103: 93

Answer

The following is the sketch of the equation.

Work Step by Step

$x^2+y^2+6x-12y+45=0$ $x^2+6x+y^2-12y=-45$ (Add both with $9$ and $36$ to complete the square) $x^2+6x+9+y^2-12y+36=-45+9+36$ $(x+3)^2+(y-6)^2=0$ The only point which satisfies the last equation is $(x,y)=(-3,6)$. So, the graph of the equation is one point $(-3,6)$.
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