Elementary Geometry for College Students (7th Edition) Clone

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: 25

Answer

$(x-1)^2+(y-2)^2+(z-3)^2 = 25$

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 $(1,2,3)$ We can find the radius of the sphere: $r = \sqrt{(5-1)^2+(5-2)^2+(3-3)^2}$ $r = \sqrt{(4)^2+(3)^2+(0)^2}$ $r = \sqrt{16+9+0}$ $r = \sqrt{25}$ The radius of the sphere is $5$ We can find an equation for the sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ $(x-1)^2+(y-2)^2+(z-3)^2 = (5)^2$ $(x-1)^2+(y-2)^2+(z-3)^2 = 25$
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