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: 48


Center: $(1,\dfrac{-1}{2},-3)$ and radius: $5$

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 $(x -1)^2+(y+\dfrac{1}{2})^2+(z +3)^2=25$ or, $(x -1)^2+(y-(-\dfrac{1}{2}))^2+(z -(-3))^2=5^2$ Compare the above equation with equation (1): Center: $(1,\dfrac{-1}{2},-3)$ and radius: $5$
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