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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 − 8 y − 6 z + 13 = 0 .

The center and radius of the sphere x 2 + y 2 + z 2 − 8 y − 6 z + 13 = 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 68E

To determine

To find: The center and radius of the sphere x2+y2+z2−8y−6z+13=0 .

Expert Solution & Answer

Answer to Problem 68E

The center and radius of the sphere x2+y2+z2−8y−6z+13=0 are (0,4,3) and 23 respectively.

Explanation of Solution

Given information:

The equation of the sphere is x2+y2+z2−8y−6z+13=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+z2−8y−6z+13=0x2+(y2−8y)+(z2−6z)+13=0x2+(y2−8y+42−42)+(z2−6z+32−32)+13=0(x−0)2+(y−4)2+(z−3)2−42−32+13=0

Further simplify,

(x−0)2+(y−4)2+(z−3)2−42−32+13=0(x−0)2+(y−4)2+(z−3)2=12(x−0)2+(y−4)2+(z−3)2=(12)2(x−0)2+(y−4)2+(z−3)2=(23)2

Compare the equation of sphere (x−0)2+(y−4)2+(z−3)2=(23)2 with general equation. It gives the value of h , k , j as 0 , 4 , 3 respectively and the value of r as 23 .

Therefore, the center and radius of the sphere x2+y2+z2−8y−6z+13=0 are (0,4,3) and 23 respectively.

Chapter 10.1, Problem 67E

Chapter 10.1, Problem 69E

Chapter 10 Solutions

Precalculus with Limits: A Graphing Approach

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