A circular disc of radius $5$ 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 + s (\cos t \ u_1 + \sin t \ u_2) $
where $s = 5$ is the radius of the disc, and $u_1, u_2$ are two mutually orthogonal vectors that are orthogonal to $n$, the unit normal vector to the plane of the 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 + s( \cos t_1 \ u_{1z} + \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 = - \sin t_1 \ u_{1z} + \cos t_1 \ u_{2z} $
Simple algebraic manipulation leads to
$z_C = s \sqrt{ u_{1z}^2 + u_{2z}^2 } $
Similar formulas apply for $x_C $ and $y_C$
Since $[u_1, u_2, n]$ forms an orthonormal set, then
$ u_{1z}^2 + u_{2z}^2 = 1 - n_z^2 $
and similarly,
$ u_{1x}^2 + u_{2x}^2 = 1 - n_x^2 $
$ u_{1y}^2 + u_{2y}^2 = 1 - n_y^2 $
Hence,
$ (x_C, y_C, z_C) = s ( \sqrt{1 - n_x^2} , \sqrt{1 - n_y^2} , \sqrt{1 - n_z^2} ) $
which leads to
$ x_C^2 + y_C^2 + z_C^2 = 2 s^2 $
Thus the center of the disc lies on a sphere centered at the origin with radius equal to $\sqrt{2} s $.
My Question:
I am looking for a confirmation of the above analysis. Alternative solutions or comments are highly appreciated. Thanks to all.
Let the plane of the disk cut coordinate axes at points $A=(x,0,0)$, $B=(0,y,0)$, $C=(0,0,z)$. The tangency condition implies that the perimeter of the disk is the incircle of triangle $ABC$ and its center is the incenter $O$ of $ABC$. From the incenter formula we then get: $$ O={aA+bB+cC\over a+b+c}, $$ where $a=\sqrt{y^2+z^2}$, $b=\sqrt{x^2+z^2}$, $c=\sqrt{x^2+y^2}$ are the lengths of the sides opposite to $A$, $B$, $C$ respectively. Squaring this equality we obtain: $$ \tag{1} O^2={a^2x^2+b^2y^2+c^2z^2\over(a+b+c)^2} =2{x^2z^2+y^2z^2+x^2y^2\over(a+b+c)^2}. $$ On the other hand, the area of triangle $ABC$ is: $$ area={1\over2}c\cdot h_c={1\over2}c\cdot\sqrt{z^2+{x^2y^2\over c^2}} ={1\over2}\sqrt{x^2z^2+y^2z^2+x^2y^2} $$
and this must be the same as inradius $s$ times half-perimeter, that is: $$ {1\over2}s(a+b+c)={1\over2}\sqrt{x^2z^2+y^2z^2+x^2y^2}. $$ Squaring both sides we get $$ s^2(a+b+c)^2=x^2z^2+y^2z^2+x^2y^2 $$ and a comparison with $(1)$ gives $$O^2=2s^2.$$