Question

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.

Answer 1

How do you find the equation if you do not know the radius or diameter?

Answer 2

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?

Answer 3

(x+1)^2+(y-3)^2+(z-2)^2=r^2

Answer 4

Now find R

Answer 5

I'm not sure how to do that...

Take the root of both sides and I'm left with r, yes?

Answer 6

Use Enhanced Pythagorean Theorem

Answer 7

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

Answer 8

x^2+y^2+z^2=r^2

Answer 9

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

Answer 10

Me and you are beast buddies @jim_thompson5910

Answer 11

Ok thanks for the help...