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High School Math Calculus Precalculus with Limits: A Graphing Approach

The center and radius of the sphere 4 x 2 + 4 y 2 + 4 z 2 − 8 x + 16 y − 1 = 0 .

The center and radius of the sphere 4 x 2 + 4 y 2 + 4 z 2 − 8 x + 16 y − 1 = 0 .

Precalculus with Limits: A Graphing Approach

Precalculus with Limits: A Graphing Approach

7th Edition

ISBN: 9781305071711

Author: Ron Larson

Publisher: Brooks/Cole

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Question

Chapter 10.1, Problem 71E

To determine

To find: The center and radius of the sphere 4x2+4y2+4z2−8x+16y−1=0 .

Expert Solution & Answer

Answer to Problem 71E

The center and radius of the sphere 4x2+4y2+4z2−8x+16y−1=0 are (1,−2,0) and 212 respectively.

Explanation of Solution

Given information:

The equation of the sphere is 4x2+4y2+4z2−8x+16y−1=0 .

Calculation:

The general standard form of the equation of sphere is (x−h)2+(y−k)2+(z−j)2=r2 with center (h,k,j) and r radius.

Simplify the given equation of sphere.

4x2+4y2+4z2−8x+16y−1=04(x2−2x)+4(y2+4y)+4z2=14(x2−2x+12−12)+4(y2+4y+22−22)+4z2=14(x−1)2+4(y+2)2+4z2−4−16=1

Further simplify,

4(x−1)2+4(y+2)2+4z2−4−16=14(x−1)2+4(y+2)2+4z2=21(x−1)2+(y+2)2+z2=214(x−1)2+(y+2)2+z2=(212)2

Compare the equation of sphere (x−1)2+(y+2)2+z2=(212)2 with general equation. It gives the value of h , k , j as 1 , −2 , 0 respectively and the value of r as 212 .

Therefore, the center and radius of the sphere 4x2+4y2+4z2−8x+16y−1=0 are (1,−2,0) and 212 respectively.

Chapter 10.1, Problem 70E

Chapter 10.1, Problem 72E

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

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