   Chapter 7.1, Problem 35E ### 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 Equation of a Sphere In Exercises 29-38, find the standard equation of the sphere with the given characteristics. See Examples 4 and 5.Endpoints of diameter: ( − 3 , 6 , 1 ) ,   ( 1 , − 5 , 2 )

To determine

To calculate: The standard equation of sphere that has points (3,6,1) and (1,5,2) as endpoints of a diameter.

Explanation

Given Information:

The endpoints of the diameter, (3,6,1) and (1,5,2).

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 the endpoints of the diameter, (3,6,1) and (1,5,2).

Since, the midpoint of the diameter of a sphere is the centre of sphere. So, by the use of midpoint formula,

Midpoint=(x1+x22,y1+y22,z1+z22)=(3+12,6+(5)2,1+22)=(1,12,32)

Thus, the centre of the sphere is (1,12,32).

Since, the distance between the centre and circumference point of the sphere is the radius of the sphere

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