# Sphere from tree points

Equation of sphere with three point (a,0,0), (0, a,0), (0,0, a) and center lies on plane x+y+z=a

Correct result:

e =  0

#### Solution:

$C: x_{0}+y_{0}+z_{0}=a \ \\ (x-x_{0})^2+(y-y_{0})^2+(z-z_{0})^2=r^2 \ \\ \ \\ (a-x_{0})^2+(0-y_{0})^2+(0-z_{0})^2=r^2 \ \\ (0-x_{0})^2+(a-y_{0})^2+(0-z_{0})^2=r^2 \ \\ (0-x_{0})^2+(0-y_{0})^2+(a-z_{0})^2=r^2 \ \\ \ \\ (a-x_{0})^2+(0-y_{0})^2+(0-a-x_{0}-y_{0})^2=r^2 \ \\ (0-x_{0})^2+(a-y_{0})^2+(0-a-x_{0}-y_{0})^2=r^2 \ \\ (0-x_{0})^2+(0-y_{0})^2+(a-a-x_{0}-y_{0})^2=r^2 \ \\ \ \\ \ \\ (a-x_{0})^2+y_{0}^2+(a+x_{0}+y_{0})^2=r^2 \ \\ x^2+(a-y_{0})^2+(a+x_{0}+y_{0})^2=r^2 \ \\ x_{0}^2+y_{0}^2+(x_{0}+y_{0})^2=r^2 \ \\ \ \\ r=\sqrt{ 6 } a \ \\ x_{0}=-a \ \\ y_{0}=-a \ \\ z_{0}=3a \ \\ \ \\ (x+a)^2 + (y+a)^2+(z-3a)^2=6 \ a^2 \ \\ \ \\ e=0$

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Math student
how can you write  r=a.(2/3)1/2
is this applicable for all plane equation?

Dr Math
We found some bugs in this problem, but I think now is OK solution:

(x+a)2 + (y+a)2+(z-3a)2 = 6 a2

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