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: 52


$(x - 0)^2+(y +1)^2+(z - 5)^2=4$

Work Step by Step

As we know the standard equation of sphere is $(x-a)^2+(y-b)^2+(z-c)^2=r^2$ ...(a) where $(a,b,c)$ represents center and radius of the sphere is $r$ Plug $a=0,b=-1,c=5, r=2$ in equation (a). Thus, we have $(x - 0)^2+(y -(-1))^2+(z - 5)^2=(2)^2$ or, $(x - 0)^2+(y +1)^2+(z - 5)^2=4$
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