University Calculus: Early Transcendentals (3rd Edition)

Published by Pearson
ISBN 10: 0321999584
ISBN 13: 978-0-32199-958-0

Chapter 11 - Section 11.1 - Three-Dimensional Coordinate Systems - Exercises - Page 600: 51


$(x - 1)^2+(y -2)^2+(z - 3)^2=14$

Work Step by Step

We know that the standard equation of a sphere is written as: $(x-x_0)^2+(y-y_0)^2+(z-z_0)^2=r^2$ ...(1) Here, $(x_0,y_0,z_0)$ represents the center and $r$ is the radius of the sphere. Since, we have $x_0=1,y_0=2,z_0=3, r=\sqrt{14}$ Plug all the above values in equation (1), then we get Thus,, $(x - 1)^2+(y -2)^2+(z - 3)^2=(\sqrt{14})^2$ Hence, $(x - 1)^2+(y -2)^2+(z - 3)^2=14$
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