No, it won't work.
The transformation of a sphere by a projection into NDC is not a sphere. You pick a single point on the sphere, and then assume you get the whole bounding volume back with just a single distance, but it is incorrect (i tried finding pictures of the resulting volume but failed. While you can see the deformation on how a sphere turns into an ellipsoid in xy plane, the transformation along the z-axis is the one that is most dramatic).
Let's walk through an example (using OpenGL projection, but the principle is the same in other NDC spaces), with near=1, far=infinity, -left=right=-bottom=top=1
. The resulting projection matrix is:
[1 0 0 0]
[0 1 0 0]
[0 0 -1 -2]
[0 0 -1 0]
Take a sphere centered at [0, 0, -10]
, of radius 1... And let's transform 3 points on the sphere:
[0, 1, -10, 1],
[0, 0, -9, 1] and
[0, 0, -11, 1]
the post-projection coords give resp:
[0, 1, 10 - 2, 10]
[0, 0, 9 - 2, 9]
[0, 0, 11 - 2, 11]
in NDC, this gives:
[0, 0.1, 0.8 ]
[0, 0, 0.777...]
[0, 0, 0.8181..]
so, the sphere points are respectively 0.1, 0.0222.., 0.01818... away from the center in NDC.
This is a single example, but you can clearly see they're not on a sphere in NDC space, and the Z direction completely changes the proportions along Z.
Not only that, but actually transforming to the NDC is incorrect if you don't clip your volume to the clipping planes first.
Say for example you want to test the point [0, 0, 2]
(behind the camera). The projection gives: [0, 0, -3, -4]
. A naïve simple w divide gives [0, 0, 0.75]
... and that's "inside" the NDC cube.