## Calculus: Early Transcendentals 8th Edition

Published by Cengage Learning

# Chapter 12 - Section 12.1 - Three-Dimensional Coordinate Systems - 12.1 Exercises: 18

#### Answer

The center is $(-4,3,-1)$ and the radius is $3$.

#### 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. $x^2+y^2+z^2+8x-6y+2z+17=0$ $(x^2+8x)+(y^2-6y)+(z^2+2z)=-17$ (rearrange terms) $(x^2+8x+16)+(y^2-6y+9)+(z^2+2z+1)=-17+16+9+1$ (complete the square) $(x^2+4)^2+(y^2-3)^2+(z^2+1)^2=9$ Now that the equation is in the form: $(x−h)^2+(y−k)^2+(z−l)^2=r^2$ The center is $(-4,3,-1)$ and the radius is $3$.

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