   Chapter 7.1, Problem 43E ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919

#### Solutions

Chapter
Section ### Calculus: An Applied Approach (Min...

10th Edition
Ron Larson
ISBN: 9781305860919
Textbook Problem
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# Finding the Center and Radius of a Sphere In Exercises 39-44, find the center and radius of the sphere. See Example 6. 2 x 2 + 2 y 2 + 2 z 2 − 4 x − 12 y − 8 z + 3 = 0

To determine

To calculate: The centre and radius of sphere whose equation is 2x2+2y2+2z24x12y8z+3=0.

Explanation

Given Information:

The equation of sphere, 2x2+2y2+2z24x12y8z+3=0.

Formula used:

The standard equation of a sphere with centre at (h,k,j) and radius r is,

(xh)2+(yk)2+(zj)2=r2

Calculation:

Consider equation of sphere,

2x2+2y2+2z24x12y8z+3=02(x2+y2+z24x6y4z+32)=0x2+y2+z24x6y4z+32=0

Convert the provided equation into the square by grouping the similar variable. So, the above equation can be written as:

(x22x)+(y26y)+(z24z)+32=0(x22x)+(y26y)+(z24z)=32

To complete the square, add the half of square of each linear term of the equation. So, to complete the square of (x22x), add [12(2)]2=1 on both sides

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