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: 56


Center: $(0,3,-4 )$ 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 $x^2+y^2+z^2-6y+8z=0$ or, $x^2+y^2-6y+z^2+8z=0$ or, $x^2+(y-3)^2+(z+4)^2=0+9+16$ or, $x^2+(y-3)^2+(z+4)^2=25$ or, $(x -0)^2+(y-3)^2+(z-(-4))^2=5$ Compare the above equation with equation (1), we get Center: $(0,3,-4 )$ and radius: $5$
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