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: 53


$(x + 1)^2+(y -\dfrac{1}{2})^2+(z+ \dfrac{2}{3})^2=\dfrac{16}{81}$

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 $a_0=-1,b=\dfrac{1}{2},c=\dfrac{-2}{3}, r=\dfrac{4}{9}$ in equation (a). Thus, we have $(x - (-1))^2+(y -(\dfrac{1}{2}))^2+(z - (\dfrac{-2}{3}))^2=(\dfrac{4}{9})^2$ or, $(x + 1)^2+(y -\dfrac{1}{2})^2+(z+ \dfrac{2}{3})^2=\dfrac{16}{81}$
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