How do I find the equation of a sphere that passes through the origin and whose center is #(4, 1, 2)#?

1 Answer
Oct 28, 2015

Answer:

The equation is: #(x-4)^2+(y-1)^2+(z-2)^2=21#

Explanation:

The general equation of a sphere with a center #C=(x_c,y_c,z_c)# and radius #r# is:

#(x-x_c)^2+(y-y_c)^2+(z-z_c)^2=r^2#

In this case center is given #C=(4,1,2)#.

To calculate the radius we use the second point given - the origin. So the radius is the distance between point #C# and the origin:

#r=sqrt((x_C-x_O)^2+(y_C-y_O)^2+(z_C-z_O)^2)=#

#=sqrt((4-0)^2+(1-0)^2+(2-0)^2)=sqrt(16+1+4)=sqrt(21)#

Now when we have all required data we can write the equation of the sphere:

#(x-4)^2+(y-1)^2+(z-2)^2=21#