Find the sphere through the circle x^2+y^2=11, z=0 that is…

Find the sphere through the circle $x^2+y^2=11,\ z=0$ that is tangent to the plane $x+y+z-5=0$.

Explanation of Method

Spheres through a circle in xOy that touch a given plane.

Step 1: Conditions

The center lies on the axis of the circle, $(0,0,c)$. Passing through the circle gives $R^2=11+c^2$. Tangency gives $R=\frac{|c-5|}{\sqrt3}$. So

\[ 3(11+c^2)=(c-5)^2\ \Longrightarrow\ c^2+5c+4=0\ \Longrightarrow\ c=-1\ \text{ or }\ c=-4 . \]

Step 2: Two spheres

\[x^2+y^2+(z+1)^2=12\qquad\text{and}\qquad x^2+y^2+(z+4)^2=27\]

Checks: $\tfrac{6}{\sqrt3}=2\sqrt3$ and $\tfrac9{\sqrt3}=3\sqrt3$. $\checkmark$

The tangent points are the feet of the perpendiculars from the centers: $(1,1,1)$ and $(2,2,-1)$.

$(0,0,-1)$ $(0,0,-4)$ $x+y+z=5$

Section by the plane through $Oz$ and the direction $\{1,1,0\}$ ($s$ measured along it). The given circle appears as the orange segment. Both spheres (centers $(0,0,-1)$, $(0,0,-4)$) contain it and touch the plane (red line).

Result

\[\boxed{\,x^2+y^2+(z+1)^2=12\qquad\text{and}\qquad x^2+y^2+(z+4)^2=27\,}\]