Thomas' Calculus 13th Edition

Published by Pearson
ISBN 10: 0-32187-896-5
ISBN 13: 978-0-32187-896-0

Chapter 12: Vectors and the Geometry of Space - Section 12.1 - Three-Dimensional Coordinate Systems - Exercises 12.1 - Page 696: 55


Center is $(-2,0,2 )$ and radius is $\sqrt 8$

Work Step by Step

As we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ where $(a,b,c)$ represents center and radius of the sphere is $r$ $x^2+y^2+z^2+4x-4z=0 \implies x^2+4x+y^2+z^2-4z=0$ $ \implies (x - 2)^2+y^2+(z - 2)^2=8$ and $(x -(- 2))^2+y^2+(z - 2)^2=(\sqrt 8)^2$ Thus, we have Center is $(-2,0,2 )$ and radius is $\sqrt 8$
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