# Chapter 10 - Section 10.6 - The Three-Dimensional Coordinate System - Exercises - Page 490: 24

$x^2+y^2+z^2 = 178$

#### Work Step by Step

We can write the general equation of a sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ where $(a,b,c)$ is the center of the sphere and $r$ is the radius The center of the sphere is $(0,0,0)$ We can find the radius of the sphere: $r = \sqrt{(3-0)^2+(12-0)^2+(-5-0)^2}$ $r = \sqrt{(3)^2+(12)^2+(-5)^2}$ $r = \sqrt{9+144+25}$ $r = \sqrt{178}$ The radius of the sphere is $\sqrt{178}$ We can find an equation for the sphere: $(x-a)^2+(y-b)^2+(z-c)^2 = r^2$ $(x-0)^2+(y-0)^2+(z-0)^2 = (\sqrt{178})^2$ $x^2+y^2+z^2 = 178$

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