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 - Page 797: 20


The center is $(0,1,2)$ and the radius is $\frac{5}{\sqrt{3}}$.

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. $3x^2+3y^2+3z^2=10+6y+12z$ $(x^2)+(y^2-2y)+(z^2-4z)=\frac{10}{3}$ (rearrange terms) $(x^2)+(y^2-2y+1)+(z^2-4z+4)=\frac{10}{3}+1+4$ (complete the square) $(x)^2+(y-1)^2+(z-2)^2=\frac{25}{3}$ Now that the equation is in the form: $(x−h)^2+(y−k)^2+(z−l)^2=r^2$ The center is $(0,1,2)$ and the radius is $\frac{5}{\sqrt{3}}$.
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