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: 19

Answer

The center is (2,0,−6) and the radius is $\sqrt {40\frac{1}{2}}.$.

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. $2x^2+2y^2+2z^2=8x-24z+1$ $(x^2−4x)+y^2+(z^2+12z)=\frac{1}{2}$ (rearrange terms) $(x^2−4x+4)+y^2+(z^2+12z+36)=\frac{1}{2}+4+36$ $(x−2)^2+y^2+(z+6)^2=40\frac{1}{2}$ Now that the equation is in the form: $(x−h)^2+(y−k)^2+(z−l)^2=r^2$ The center is (2,0,−6) and the radius is $\sqrt {40\frac{1}{2}}.$
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