Babyslapmafro
  • Babyslapmafro
Please help, I am asked to find the standard equation of the sphere that satisfies the stated conditions: Center (-1,3,2) and passing through the origin.
Mathematics
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SOLVED
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chestercat
  • chestercat
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Babyslapmafro
  • Babyslapmafro
How do you find the equation if you do not know the radius or diameter?
anonymous
  • anonymous
Well, you have the general form of an sphere (x-1)^2+(y-b)^2+(z-c)^2=r^2 How are you going to find R? What are a, b, and c?
Babyslapmafro
  • Babyslapmafro
(x+1)^2+(y-3)^2+(z-2)^2=r^2

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anonymous
  • anonymous
Now find R
Babyslapmafro
  • Babyslapmafro
I'm not sure how to do that... Take the root of both sides and I'm left with r, yes?
anonymous
  • anonymous
Use Enhanced Pythagorean Theorem
jim_thompson5910
  • jim_thompson5910
using the distance formula (in 3D space) r = sqrt( (x1-x2)^2 + (y1-y2)^2 + (z1-z2)^2 ) r = sqrt( (0-x2)^2 + (0-y2)^2 + (0-z2)^2 ) ... plug in (x1,y1,z1) = (0,0,0) r = sqrt( (0-(-1))^2 + (0-3)^2 + (0-2)^2 ) ... plug in (x2,y2,z2) = (-1,3,2) I'll let you finish up
anonymous
  • anonymous
x^2+y^2+z^2=r^2
jim_thompson5910
  • jim_thompson5910
alternatively, you can plug in (0,0,0) into (x+1)^2+(y-3)^2+(z-2)^2=r^2 to get (x+1)^2+(y-3)^2+(z-2)^2=r^2 (0+1)^2+(0-3)^2+(0-2)^2=r^2 then solve for r to get the radius
anonymous
  • anonymous
Me and you are beast buddies @jim_thompson5910
Babyslapmafro
  • Babyslapmafro
Ok thanks for the help...

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