   Chapter 7.1, Problem 36E ### 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 a diameter: (2, 9, 11), (2, -3, -6)

To determine

To calculate: The standard equation of sphere that has points (2,9,11) and (2,3,6) as endpoints of a diameter.

Explanation

Given Information:

The endpoints of the diameter, (2,9,11) and (2,3,6).

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, (2,9,11) and (2,3,6).

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)=(2+22,9+(3)2,11+(6)2)=(2,3,52)

Thus, the centre of the sphere is (2,3,52).

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

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