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Precalculus Enhanced with Graphing Utilities

Sullivan

$Math Studyset 91影视 Explanations$ Math

6th Edition 路 1200 pages

ISBN: 9780321795465

Chapter 9: Q. 76 (page 622)

In Problems $71 鈥� 76$ , find the radius and center of each sphere.
$3 x^{2} + 3 y^{2} + 3 z^{2} + 6 x - 6 y = 3$

Short Answer

Expert verified

Radius is $\sqrt{3}$ and center is $(- 1, 1, 0)$

Step by step solution

01

Step 1. Given information

An equation is given as $3 x^{2} + 3 y^{2} + 3 z^{2} + 6 x - 6 y = 3$

02

Step 2. Finding radius and center

Convert the equation in the form

$r^{2} = {(x - x_{0})}^{2} + {(y - y_{0})}^{2} + {(z - z_{0})}^{2}$

Where r is radius and center is $P_{0} = (x_{0}, y_{0}, z_{0})$

Simplify,

$3 x^{2} + 3 y^{2} + 3 z^{2} + 6 x - 6 y = 3 x^{2} + y^{2} + z^{2} + 2 x - 2 y = 1 (x^{2} + 2 x) + (y^{2} - 2 y) + z^{2} = 1 (x^{2} + 2 x + 1) + (y^{2} - 2 y + 1) + z^{2} = 1 + 1 + 1 (x + 1)^{2} + (y - 1)^{2} + z^{2} = 3$

Therefore the center is $(- 1, 1, 0)$ and radius is $\sqrt{3}$

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