Answer
a) $(x-2)^{2}+(y+3)^{2}+(z-6)^{2}=36$
b) $(x-2)^{2}+(y+3)^{2}+(z-6)^{2}=4$
c) $(x-2)^{2}+(y+3)^{2}+(z-6)^{2}=9$
Work Step by Step
The center: $C(2, -3, 6)$
The equation of a sphere is $(x-h)^{2}+(y-k)^{2}+(z-l)^{2}=r^{2}$
a) Since the sphere touches the xy plane, we know that the value of $z=0$, this means that the radius of the sphere will be the value of the center's $z$ coordinate. Therefore, the radius of the sphere is $r=6$ and $r^{2}=36$.
The equation of the sphere is: $$(x-2)^{2}+(y+3)^{2}+(z-6)^{2}=36$$
b) Since the sphere touches the yz plane, we know that the value of $x=0$, this means that the radius of the sphere will be the value of the center's $x$ coordinate. Therefore, the radius of the sphere is $r=2$ and $r^{2}=4$.
The equation of the sphere is: $$(x-2)^{2}+(y+3)^{2}+(z-6)^{2}=4$$
c) Since the sphere touches the xz plane, we know that the value of $y=0$, this means that the radius of the sphere will be the value of the center's $y$ coordinate. Therefore, the radius of the sphere is $r=3$ and $r^{2}=9$.
The equation of the sphere is: $$(x-2)^{2}+(y+3)^{2}+(z-6)^{2}=9$$
