Find the Equation of a Sphere with Given Endpoints and Midpoint

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Homework Help Overview

The problem involves finding the equation of a sphere given the endpoints of a diameter and requires calculating the midpoint and radius based on those points.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants discuss how to find the midpoint of the endpoints and question the calculations related to the diameter and radius of the sphere.

Discussion Status

Some participants have provided insights into the midpoint formula and the calculation of the diameter's length, while others are questioning the accuracy of the calculations presented. There is an ongoing exploration of the correct values for the midpoint and radius.

Contextual Notes

There appears to be some confusion regarding the calculations of the diameter and radius, with differing interpretations of the results. The original poster expresses uncertainty about the midpoint calculation.

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Homework Statement



Find an equation of the sphere one of whose diameters has (1, 3, -5) and
(3, -1, 3) as its endpoints.

Homework Equations



midpoint.

The Attempt at a Solution



i don't understand how to find the midpoint, even though its the mean of:
(2, 1, 1)

i've also been given the hint to find the measurement of a great circle:
sqrt( 4 + 16 + 64 ) = sqrt(80)


which equals: Thus,
(x-2) ^ 2 + (y-1) ^ 2 + (z-1) ^ 2 = 80

please help me solve this problem, i almost got it!
 
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I think the center's at (2,1,-1) and from here it follows that the radius is [itex]\sqrt{21}[/itex]
 
sqrt( 4 + 16 + 64 ) = sqrt(80)?

Shouldn't it be sqrt( 4 + 16 + 64 ) = sqrt(84)
 
The midpoint of the line segment from (x0,y0,z0) to (x1,y1,z1) is
[tex]\left(\frac{x_0+x_1}{2},\frac{y_0+y_1}{2},\frac{z_0+z_1}{2}\right)[/tex]
I thought everyone knew that!
The length of the diameter is [itex]\sqrt{84}= 2\sqrt{21}[/itex].
The length of the radius is half that: [itex]\sqrt{21}[/itex]
 

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