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: 16

Answer

$(x−1)^2+(y−2)^2+(z−3)^2=14$

Work Step by Step

First, determine the radius of the sphere using the distance formula: $d=\sqrt{(x2−x1)^2+(y2−y1)^2+(z2−z1)^2}$ $r=\sqrt{(1-0)^2+(2-0)^2+(3-0)^2}=\sqrt{14}$ 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. $(x−1)^2+(y−2)^2+(z−3)^2=14$
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