Write an equation for a sphere passing through a circle and tangent to a plane I'm trying to solve this task:
'''Write an equation for a sphere passing through a circle $x^2 + y^2 = 11$ and tangent to a plane $x + y + z - 5 = 0$.'''
Center of the sphere should moves only in axis Z, so it has coordinates $(0, 0, \alpha)$. I also found that
$\alpha^2 = R^2 - 11$, where R - radius of a sphere.
I came to this system:

*

*$x_0^2 + y_0^2 + (z_0 - \sqrt(R^2 - 11))^2 = R^2$

*$x_0 + y_0 + z_0 = 5$
where $x_0, y_0, z_0$ are coordinates of a touch point of a sphere with the plane.
The correct answer is two spheres:
$x^2 + y^2 + (z + 1)^2 = 12$
$x^2 + y^2 + (z + 4)^2 = 27$
I can't figure out how to find these two values of the parameter $\alpha$. Could somebody please explain how to do it?
Thanks in advance.

If the center of the sphere is $(x_0, y_0, z_0)$, and its radius is $R$ then
its equation is
$ (x - x_0)^2 + (y - y_0)^2 + (z - z_0)^2 = R^2 $
Set $z = 0 $ , then you get
$ (x - x_0)^2 + (y - y_0)^2 + z_0^2 = R^2 $
Comparing this with $ x^2 + y^2 = 11 $ we deduce that $x_0 = y_0 = 0 $ and
$ 11 = R^2 - z_0^2 $
Since the sphere is tangent to $ x+ y+z = 5 $, then the distance of the center which is (0, 0, z_0) to the plane is equal to $R$ , thus
$ R = \dfrac{| 0 + 0 + z_0 - 5 |} {\sqrt{3}} $
Squaring
$ R^2 = \dfrac{(z_0 - 5)^2}{3} $
Thus
$ 11 + z_0^2 = \dfrac{(z_0 - 5)^2 }{3} $
So that,
$ 33 + 3 z_0^2 = z_0^2 - 10 z_0 + 25 $
From which,
$ 2 z_0^2 + 10 z_0 + 8 = 0 $
Dividing through by $2$,
$ z_0^2 + 5 z_0 + 4 = 0 $
Factoring,
$ (z_0 + 4)(z_0 + 1) = 0 $
Therefore, $z_0 = -4$ or $z_0 = -1$, and corresponding to that,
$R^2 =  27 $ or $ R^2 = 12 $
And the equations of the two spheres are
$ x^2 + y^2 + (z + 4)^2 = 27 $
and
$ x^2 + y^2 + (z + 1)^2 = 12 $
