Answer:
[tex](x-3)^2+(y+2)^2+(z-8)^2=64[/tex]
Step-by-step explanation:
We are given that a sphere of radius 8 that lies above the xy-plane and a sphere just touches the xy-plane at the point (3,-2,0).
We have to find the equation of sphere
We know that distance between the center (h,k,l)of sphere and the point (3,-2,0) lies on the sphere is equal to radius of sphere.
[tex]8=\sqrt{(3-h)^2+(-2-k)^2+(-l)^2}[/tex]
Therefore, the center of sphere is at(3,-2,8)
Equation of sphere with center (h,k,l) and radius r
[tex](x-h)^2+(y-k)^2+(z-l)^2=r^2[/tex]
Substitute the values then we get
Hence,the required equation of sphere
[tex](x-3)^2+(y+2)^2+(z-8)^2=64[/tex]
