PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)
8th Edition
ISBN: 9781337904285
Author: Larson
Publisher: CENGAGE L
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Chapter 10, Problem 12CR
To determine

To find: the standard equation of sphere.

Expert Solution & Answer
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Answer to Problem 12CR

The standard equation of sphere is (x+2)2+(y−6)2+(z+1)2=11 .

Explanation of Solution

Given information:

Endpoints of a diameter: (−3,5,−2),(−1,7,4) .

Formula used: the equation of the sphere with centre and radius,

  (x−h)2+(y−k)2+(z−j)2=r2

the distance formula,

  d=(x2−x1)2+(y2−y1)2+(z2−z1)2

the midpoint formula,

  m=(x1+x22,y1+y22,z1+z22)

Calculation: since, endpoints of diameter are given, the centre of the sphere will be the midpoint of the endpoints of diameter, so to find the centre first substitute in the midpoint formula and then simplify,m=(x1+x22,y1+y22,z1+z22)m=(−3−12,5+72,−2+42)     [substitute]m=(−2,6,−1)                              [simplify]

The radius of the sphere is the half of the distance between of the endpoints of diameter,

  d=(x2−x1)2+(y2−y1)2+(z2−z1)2d=(−1+3)2+(7−5)2+(4+2)2        [substitute]d=4+4+36                                          [subtract,add and square]d=44                                           [add]d=211

Thus, the radius of sphere is r=11 , to find the equation of sphere, first substitute and then simplify,

  (x−h)2+(y−k)2+(z−j)2=r2(x+2)2+(y−6)2+(z+1)2=11   [substitute]

Thus, The standard equation of sphere is (x+2)2+(y−6)2+(z+1)2=11 .

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Chapter 10, Problem 11CR
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Chapter 10, Problem 13CR

Chapter 10 Solutions

PRECALCULUS W/LIMITS:GRAPH.APPROACH(HS)

Ch. 10.1 - Prob. 1ECh. 10.1 - Prob. 2ECh. 10.1 - Prob. 3ECh. 10.1 - Prob. 4ECh. 10.1 - Prob. 5ECh. 10.1 - Prob. 6ECh. 10.1 - Prob. 7ECh. 10.1 - Prob. 8ECh. 10.1 - Prob. 9ECh. 10.1 - Prob. 10E
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