The intersection of a sphere and the xy-plane is the circle $(x-1)^2+(y-2)^2=5$ and point $(1,2,1)$ is on the sphere The intersection of a sphere and the $xy$-plane is the circle $$(x-1)^2+(y-2)^2=5$$ and point $(1,2,1)$ is on the sphere. Find the center and radius of the sphere.
When the sphere and $xy$-plane intersect than $z=0$. 
I tried $(x-1)^2+(y-2)^2+(z+?)^2=5$ but what can I do next? 
 A: Hint Note that the point $(1,2,1)$ is directly above the center of the circle. This means it is the top of the sphere. The center of the sphere is therefore a point $(1,2,a)$ equidistant from the points of the circle and the point $(1,2,1).$
A: Just put $(x,y,z)=(1,2,1)$ in the equation $(x-1)^{2}+(y-2)^{2}+(z-c)^{2}=c^{2}+5$. You get $(1-c)^{2}=5$ so $c=-2$. The center is $(1,2,-2)$ and radius is $\sqrt 9=3$. 
A: Hint: take any point $A$ on that circle and calculate the plane which perpendicular bisector of segment $AB$ where $B=(1,2,1)$. Point where this plane cuts line $(1,2,z)$ is center.
A: Let $O(a,b,c)$ be center of sphere, $A(1,2,0)$ be the center of intersection circle, $B(1,2+\sqrt{5},0)$ be the point of the sphere and intersection circle, $C(1,2,1)$ be the point on the sphere.
Then the radius of the sphere is $R=OC=OB=AO+AC=AO+1$. 
The triangle $AOB$ is right, so:
$$AO^2+AB^2=OB^2 \Rightarrow AO^2+5=(AO+1)^2 \Rightarrow AO=2.$$
Thus:
$$R=AO+1=3;O(1,2,-2).$$
A: One more :
Equation of the sphere:
$(x-a)^2+(y-b)^2+(z-c)^2=R^2.$
Cut with plane $z=0:$
$(x-a)^2+(y-b)^2 +c^2=R^2.$
1) Compare with the given circle:
$a=1,b=2, R^2-c^2= (√5)^2$
2) $(1,2,1)$ lies on the sphere:
$0+0+(1-c)^2=R^2.$
Hence: $1-2c+c^2=R^2;$
$1-2c= R^2-c^2=(√5)^2.$
$c= (1-5)/2=-2$, and $R^2 =9$.
