An elliptical disc of semi-major axis $a$ and semi-minor axis $b$ is placed in the corner of the first octant (where $x \ge 0, y \ge 0, z \ge 0$), such that it is tangent to the $xy, xz$ and $yz$ planes. Find the locus of the center of the disc as its normal vector sweeps all possible orientations.
My Attempt:
Points on the perimeter of the disc can be parameterized as
$ r(t) = r_C + a \cos t \ u_1 + b \sin t \ u_2 $
where $u_1, u_2$ are two mutually orthogonal unit vectors that are orthogonal to $n$, the unit normal vector to the plane of the disc. $u_1$ is along the major axis and $u_2$ is along the minor axis of the elliptical disc. And,
$ r_C = (x_C, y_C, z_C) $
is the center of the disc.
At tangency with the $xy$ plane, we have
$ 0 = z_C + a \cos t_1 \ u_{1z} + b \sin t_1 \ u_{2z} $
For some $t_1 \in [0, 2 \pi) $.
And by setting the $z$ coordinate of the tangent vector to $r$ (which is $r'$) to $0$, we get
$ 0 = - a \sin t_1 \ u_{1z} + b \cos t_1 \ u_{2z} $
Simple algebraic manipulation leads to
$z_C = \sqrt{ a^2 u_{1z}^2 + b^2 u_{2z}^2 } $
Similar formulas apply for $x_C $ and $y_C$, namely,
$x_C = \sqrt{ a^2 u_{1x}^2 + b^2 u_{2x}^2 } $
$y_C = \sqrt{ a^2 u_{1y}^2 + b^2 u_{2y}^2 } $
Squaring $x_C, y_C , z_C $ and adding yields
$ x_C^2 + y_C^2 + z_C^2 = a^2 (u_{1x}^2 + u_{1y}^2 + u_{1z}^2) + b^2 (u_{2x}^2 + u_{2y}^2 + u_{2z}^2 ) $
But $u_1$ and $u_2$ are unit vectors, therefore, the above simplifies to
$ x_C^2 + y_C^2 + z_C^2 = a^2 + b^2 $
Thus the center of the disc lies on a sphere centered at the origin with radius equal to $\sqrt{a^2 + b^2} $.
My Question:
I am looking for a confirmation of the above analysis. Alternative solutions or comments are highly appreciated. Thanks to all.