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: 17

Answer

The center is $(1,2,-4)$ and the radius is $6$.

Work Step by Step

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. To get the given equation into this form, we must complete the square for all three variables. In order to do this add $(\frac{b}{2})^2$ to both sides of the equation ($b$ is the coefficient before the $x$, $y$, or $z$ term). Since there are three different variables, we must complete the square three times. $x^2+y^2+z^2-2x-4y+8z=15$ $(x^2-2x)+(y^2-4y)+(z^2+8z)=15$ (rearrange terms) $(x^2-2x+1)+(y^2-4y+4)+(z^2+8z+16)=15+1+4+16$ $(x^2-1)^2+(y^2-2)^2+(z^2+4)^2=36$ Now that the equation is in the form: $(x−h)^2+(y−k)^2+(z−l)^2=r^2$ The center is $(1,2,-4)$ and the radius is $6$.
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