Calculus 8th Edition

Published by Cengage
ISBN 10: 1285740629
ISBN 13: 978-1-28574-062-1

Chapter 12 - Vectors and the Geometry of Space - 12.1 Three-Dimensional Coordinate Systems - 12.1 Exercises: 15

Answer

$(x-3)^2+(y-8)^2+(z-1)^2=30$

Work Step by Step

First we find the radius of the sphere by finding the distance from the center to one of the points: $r=\sqrt{(3-4)^2+(8-3)^2+(1--1)^2}=\sqrt{(-1)^2+5^2+2^2}=\sqrt{1+25+4}=\sqrt{30}$ Next we use the equation of the sphere with the center point and radius: $(x-3)^2+(y-8)^2+(z-1)^2=(\sqrt{30})^2=30$
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