Find an equation of the sphere with center $( 6 , - 3,6 )$ that touches the xy-plane.
A) $( x + 6 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 36$
B) $( x - 6 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 36$
C) $( x + 6 ) ^ { 2 } + ( y - 3 ) ^ { 2 } + ( z + 6 ) ^ { 2 } = 36$
D) $( x - 6 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 6$
E) $( x - 6 ) ^ { 2 } + ( y + 3 ) ^ { 2 } + ( z - 6 ) ^ { 2 } = 40$

Question

Quizplus · Accepted Answer

Since the sphere touches the xy-plane, its center lies on a line perpendicular to the xy-plane passing through the point where it touches the plane. Therefore, the center of the sphere has a z-coordinate of 0. This tells us that the equation of the sphere should have a term of the form $(z-a)^2$ where *a* is the z-coordinate of the center. Plugging in the given center point $(6,-3,6)$, we get the equation $(x-6)^2 + (y+3)^2 + (z-6)^2 = r^2$. To find *r*, we note that the sphere touches the xy-plane, which means that its radius is simply the z-coordinate of its center. Therefore, *r* = 6. Substituting this value of *r* into our equation gives $(x-6)^2 + (y+3)^2 + (z-6)^2 = 36$, which matches option B.

Find an Equation of the Sphere with Center (6,−3,6)( 6 , - 3,6 )(6,−3,6)

Find an Equation of the Sphere with Center $( 6 , - 3,6 )$