Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning
ISBN 10: 1285741552
ISBN 13: 978-1-28574-155-0

Chapter 12 - Section 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, determine the radius of the sphere using the distance formula: $d=\sqrt{(x_2-x_1)^2+(y_2-y_1)^2+(z_2-z_1)^2}$ $r=\sqrt{(3-4)^2+(8-3)^2+(1+1)^2}=\sqrt{30}$ The equation for a sphere is represented by: $(x-h)^2+(y-k)^2+(z-l)^2=r^2$ in which the point $(h,k,l)$ is the center of the sphere and $r$ is the radius. Plug in the values for the center and the radius. $(x-3)^2+(y-8)^2+(z-1)^2=30$
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