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High School Math Calculus Precalculus with Limits: A Graphing Approach

The center and radius of the sphere x 2 + y 2 + z 2 + 2 y − 14 z = 0 .

The center and radius of the sphere x 2 + y 2 + z 2 + 2 y − 14 z = 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 66E

To determine

To find: The center and radius of the sphere x2+y2+z2+2y−14z=0 .

Expert Solution & Answer

Answer to Problem 66E

The center and radius of the sphere x2+y2+z2+2y−14z=0 are (0,−1,7) and 52 respectively.

Explanation of Solution

Given information:

The equation of the sphere is x2+y2+z2+2y−14z=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.

x2+y2+z2+2y−14z=0x2+(y2+2y)+(z2−14z)=0x2+(y2+2y+12−12)+(z2−14z+72−72)=0x2+(y+1)2+(z−7)2−12−72=0

Further simplify,

x2+(y+1)2+(z−7)2−12−72=0x2+(y+1)2+(z−7)2=50x2+(y+1)2+(z−7)2=(50)2x2+(y+1)2+(z−7)2=(52)2

Compare the equation of sphere x2+(y+1)2+(z−7)2=(52)2 with general equation. It gives the value of h , k , j as 0 , −1 , 7 respectively and the value of r as 52 .

Therefore, the center and radius of the sphere x2+y2+z2+2y−14z=0 are (0,−1,7) and 52 respectively.

Chapter 10.1, Problem 65E

Chapter 10.1, Problem 67E

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

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