what is the image of the disk? Consider two infinite opaque walls(planes) A and B separated by a distance c; A light emitting disk of radius r is placed parallel to the floor at a distance d from the wall B. Now a hole is punched at a height of h from the floor. What shape is the image of the disk formed on wall A?

*this crude image shows the setup

I tried to write the "process" as a complex function.
I got $f(x,y) = \frac{xc}{y-h} +i\frac{yh}{y-c}$
 where x and y is the position of a point on the floor with respect to A...
I still can't find a reasonable expression for the image.
 A: HINT:
Place the center of coordinate system at the position of the hole. Now the horizontal plane has equation
$$H \colon z = - h$$
In this plane we have a circle of radius $r$ with center at the point $(r + d, 0, - h)$ ( if $d$ were $0$, then the circle touches the wall; in general it is at the distance $d$ to the right). The equation of the circle in this horizontal plane is
$$f(x,y) \colon = (x-(d+r))^2 + y^2 - r^2=0, \ \ z = -h$$
The equation of the cone determined by this curve is homogenous in $x$, $y$, $z$ ( that is, depends only on the ratios $\frac{x}{z}$, $\frac{y}{z}$, or $[x\colon y \colon z]$, and can be easily checked to be
$$f(\frac{-h x}{z}  , \frac{-h y}{z} ) = 0$$
Now to find the intersection of the cone with the vertical plane $x=-c$, plug in $x=-c$ in the above equation in $x$, $y$, $z$, and get an equation in $y$, $z$
$$f(\frac{-h (-c)}{z}, \frac{-h y}{z}) = 0$$
$\bf{Added:}$ Let's consider a numerical case: $h= 2$, $c = 3$, $d=1$, $r = 4$.
The equation of the circle in the horizontal plane $z= -2$ is
$$(x-(1+4))^2 + y^2 - 16 = 0$$
or
$$x^2 - 10 x + y^2 + 9 = 0$$
Now homogenize ( $x \mapsto \frac{-2 x}{z}$, $y\mapsto \frac{-2y}{z}$) and get the equation of the cone
$$4 x^2 + 20 x z + 4 y^2 + 9 z^2 = 0$$
Now plug in $x=- 3$ in the above.
$\bf{Added:}$ Explanation for the equation of the cone:  along a ray of the cone all of the points  have proportional coordinates. Hence the equation of the cone has to be of the form $F(\frac{x}{z}, \frac{y}{z})= 0$. Now the idea is to obtain an equation of this form from the equation $f(x,y)=0, \ \ z=-h$.  Substitute  $x$ with $\frac{-h}{z} \cdot x$ ( note that the fraction is $1$ on the plane $z=-h$), and similarly for $y$.
