Intermediate Algebra: Connecting Concepts through Application

Published by Brooks Cole
ISBN 10: 0-53449-636-9
ISBN 13: 978-0-53449-636-4

Chapter 9 - Conic Sections, Sequences, and Series - 9.1 Parabolas and Circles - 9.1 Exercises: 63

Answer

$x^2+8x+y^2-4y+11=0$

Work Step by Step

Center=$(-4,2)$ Radius=$3$ Equation: $(x+4)^2+(y-2)^2=3^2$ $\implies x^2+8x+16+y^2-4y+4=9$ $\implies x^2+8x+y^2-4y+11=0$
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