Vtyjkyuk

When sunlight casts a shadow on an axi-symmetric inside surface we have a $ plane$ determined by rim boundary $B$ projection $E$ and mid centers $C$ on rim that divides illuminated (white) and dark portions (gray is shadow of the bowl itself below $BC$ and yellow is shadow from semicircle radius $CB$) as shown.

One can satisfy this with a parabolic dish of meridian

$$ z = k,r^2$$

How can we formulate this in general to include all meridians producing such shadow partition by a plane ?

This is not an answer, but a report of my findings. I am hoping that someone better at math can use these to actually answer the question.

Let's use a coordinate system where the base of the dish is at origin, the height of the dish is $h$, the radius of the rim is $r$. Let all light rays be parallel, with direction vector $(-1 , 0 , -m)$, $0 lt m in mathbbR$. The light rays on the terminator cast by the rim of the dish fulfill
$$z_textterminator(x, y) = m left(x - sqrtr^2 - y^2 right) + h tag1labelNA1$$
Note that $eqrefNA1$ applies to all dishes that have a circular rim of radius $r$ at $z = h$.

A parabolic dish with base at origin, height $h$, and rim radius $r$, can be defined by
$$z_textdish(x, y) = frachr^2 left( x^2 + y^2 right) tag2labelNA2$$
The light rays on the terminator cast by the dish rim intersect with the inside of the bowl when $0 le z_textterminator(x, y) = z_textdish(x, y) lt h$. Solving that for $x$ yields
$$x_textterminator(y) = fracm r^2 pm leftlvert m r^2 - 2 h sqrtr^2 - y^2 rightrvert2 h tag3labelNA3$$
To locate the $y$ coordinate where the terminator intersects the rim, we substitute $eqrefNA3$ into $z_textdish(x,y) = h$, and solve for $y$. The result is
$$y_textrim = pm r sqrt1 - fracm^2 r^24 h^2 tag4alabelNA4a$$
From $z_textterminator(x, y_textrim) = h$ we can solve the corresponding $x$ coordinate:
$$x_textrim = fracm r^22 h tag4blabelNA4b$$
and obviously $z_textrim = h$.

Note that $(x_textrim, y_textrim, z_textrim)$ is not the point $C$ in OP's diagram; it is to the right of it. Here is an orthographic projection of $h = 0.5$, $r = 1$, $m = 0.2$ I raytraced in POVRay:
Note that light comes from right, and the dish is an indentation in the cylinder.

If the slope of the light is small enough, $m lt frac2 hr$, the absolute value term in $eqrefNA3$ is negative or zero for $-lvert y_textrim rvert lt y lt lvert y_textrim rvert$. Then,
$$leftlbracebeginaligned
x_textterminator(y) &= fracm r^2h - sqrtr^2 - y^2 \
y_textterminator(y) &= y \
z_textterminator(y) &= fracm^2 r^2h - 2 m sqrtr^2 - y^2 + h \
endalignedright., quad lvert y rvert lt lvert y_textrim rvert, ; m lt frac2 hr tag4clabelNA4c$$
where $z_textterminator(y)$ was obtained by substituting $x_textterminator(y)$ into $eqrefNA1$ or $eqrefNA2$ (both yield the same result).

We can test for planarity (via torsion), using the first, second, and third derivatives. The curve is planar if and only if
$$det left [ beginmatrix
fractsqrtr^2 - t^2 & fracr^2(r^2 - t^2)^3/2 & frac3 t r^2(r^2 - t^2)^5/2 \
1 & 0 & 0 \
frac2 m tsqrtr^2 - t^2 & frac2 m r^2(r^2 - t^2)^3/2 & frac6 m t r^2(r^2 - t^2)^5/2 \
endmatrix right ] = 0$$
This is indeed true for the parabolic dish.

Unfortunately, I don't see how to generalize this to other dish profiles, not even to
$$z_textdish = frachr^2 kleft( x^2 + y^2 right)^k$$
except for $k = 1$ (the above parabolic dish case), and maybe $k = 1/2$. The issue is solving $z_textterminator(x,y) = z_textdish(x,y)$ for $x$ (to get $x_textterminator(y)$).

Let's consider a right circular cylinder with radius $r$, axis along $z$ axis, and top at $z = h$. In this case,
$$x_textcylinder(y) = pmsqrtr^2 - y^2$$
and solving $eqrefNA1$ for $x$ yields
$$x_textterminator(y, z) = sqrtr^2 - y^2 + fracz - hm$$
Solving $x_textcylinder(y) = x_textterminator(y, z)$ for $z$ yields
$$z_textterminator(y) = h - 2 m sqrtr^2 - y^2$$
(and $z_textterminator(y) = h$, but that is on the rim).
i.e. the terminator curve is
$$leftlbracebeginaligned
x_textterminator(t) &= -sqrtr^2 - t^2 \
y_textterminator(t) &= t \
z_textterminator(t) &= h - 2 m sqrtr^2 - t^2 \
endalignedright .$$
where $x_textterminator(t)$ was obtained by solving $z_textterminator(y) = z_textterminator(x, y)$ and solving for $x$.
The torsion test (determinant of the matrix formed by the first three derivatives of the terminator curve) is again
$$det left [ beginmatrix
fractsqrtr^2 - t^2 & fracr^2(r^2 - t^2)^3/2 & frac3 t r^2(r^2 - t^2)^5/2 \
1 & 0 & 0 \
frac2 m tsqrtr^2 - t^2 & frac2 m r^2(r^2 - t^2)^3/2 & frac6 m t r^2(r^2 - t^2)^5/2 \
endmatrix right ] = 0$$
Its determinant is (still!) zero, so the terminator cast into a right circular cylinder by a flat rim perpendicular to the axis of the cylinder, is also planar.