Elementary Geometry for College Students (7th Edition)

Published by Cengage
ISBN 10: 978-1-337-61408-5
ISBN 13: 978-1-33761-408-5

Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 490: 26


$(x+1)^2+(y-2)^2+(z-4)^2 = 49$

Work Step by Step

We can write the general equation of a sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ where $(a,b,c)$ is the center of the sphere and $r$ is the radius The center of the sphere is $(-1,2,4)$ The radius of the sphere is $7$ We can find an equation for the sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ $(x-(-1))^2+(y-2)^2+(z-4)^2 = (7)^2$ $(x+1)^2+(y-2)^2+(z-4)^2 = 49$
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