   Chapter 7.5, Problem 5QY ### 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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# In Exercises 4 and 5, find the standard equation of the sphere with the given characteristics.Endpoints of a diameter: (0, 3, 1), (2, 5, - 5)

To determine

To calculate: The standard form of equation of the sphere where the endpoints of a diameter of sphere is (0,3,1),(2,5,5).

Explanation

Given Information:

The endpoints of a diameter of sphere are (0,3,1),(2,5,5).

Formula used:

The standard form of a sphere is,

(xa)2+(yb)2+(zc)2=r2

Where, (a,b,c) is centre of sphere and r is radius of sphere.

Distance formula to find distance by two given points (x,y,z)and(x2,y2,z2)

d=(x1x2)2+(y1y2)2+(z1z2)2

Calculation:

Consider the standard equation,

(xa)2+(yb)2+(zc)2=r2

The endpoint of the diameter are (0,3,1),(2,5,5). Thus, the centre of the diameter is the midpoints of these points.

Therefore, the centre point is,

(0+2)2=1(3+52)=4(152)=2

Thus, the centre of the sphere is, (1,4,2)

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