Precalculus: Mathematics for Calculus, 7th Edition

Published by Brooks Cole
ISBN 10: 1305071751
ISBN 13: 978-1-30507-175-9

Chapter 9 - Section 9.3 - Three-Dimensional Coordinate Geometry - 9.3 Exercises - Page 652: 14

Answer

$(x+10)^2+(y)^2+(z-1)^2=11$

Work Step by Step

To solve this problem let us look at the formula given. We simply have to plug in our coordinates of our center and the value of the radius in place of a,b,c, and r. r is $\sqrt 11$ and $(a,b,c)$ is $(-10, 0,1)$. Plugging that in we get $(x+10)^2+(y)^2+(z-1)^2=11$
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