How to define a perspective circle in xy?

Question

You can see a perspective view of a square(FCED) and a circle in 2D screen. O is center of the circle.

How can I define the perspective circle equation that shown as red in the picture?

Thanks a lot

Answer 1

The red curve is an ellipse, but the point $O$ will not be the center of this ellipse!

The values $a$, $c$, $d$ seem to be given, and the figure is symmetric with respect to the $y$-axis. Intersecting $A\vee D$ with $B\vee C$ gives you the point $F=(-f,-g)$. It follows that $E=(f, -g)$ and $O=(0,-g)$. Intersecting $A\vee D$ with $B\vee O$ then furnishes $H$, and intersecting $B\vee C$ with $A\vee O$ furnishes $G$.

By symmetry the ellipse has an equation of the form $${x^2\over p^2} +{(y+m)^2\over q^2}=1$$ with unknown constants $p$, $q$, $m$. The coordinates of $G$ and $H$, through which the ellipse passes, provide more than enough information to determine these constants.

Answer 2

One possibility is to find out the projective transformation itself, then apply on the circle.

Were these objects originally really a square and a circle somewhere? I mean, can you use those data or only $a,c,d$?

Put the whole picture in 3d, on the $z=1$ plane (that is, $A=[-a,0,1]$, $B=[a,0,1]$, $C=[0,-c,1]$ and $D=[0,-d,1]$ will be). Then consider the matrix $$M:=[A|B|C]=\begin{bmatrix} -a & a & 0 \\ 0 & 0 & -c \\ 1 & 1 & 1 \end{bmatrix}$$ then $M\cdot\begin{bmatrix} 1\\0\\0 \end{bmatrix} = A$, $\ M\cdot\begin{bmatrix} 0\\1\\0 \end{bmatrix} = B$, $\ M\cdot\begin{bmatrix} 0\\0\\1 \end{bmatrix} = C$, so $M^{-1}$ takes $A$ and $B$ to the ideal points $i_x:=\begin{bmatrix} 1\\0\\0 \end{bmatrix}$ and $i_y:=\begin{bmatrix} 0\\1\\0 \end{bmatrix}$, and $C$ to the origo $\begin{bmatrix} 0\\0\\1 \end{bmatrix}$.

So, for this, you need to calculate $M^{-1}$, then $D':=M^{-1}D$ (divided by its 3rd coordinate) will give the opposite corner of the square, suppose $D'=\begin{bmatrix} u\\v\\w \end{bmatrix}$, then $u=v$ should hold (because it wants to be a square), and $F'=(u/w,0)$ and $E'=(0,v/w)$ will be [projected back to the $z=1$ plane].

Then form the circle (centered at $(u/2w,u/2w)$ with radius $u/2w$) in homogenious coordinates [-- so we could also call it a cone]: $$(\frac xz-\frac u{2w})^2 + (\frac yz-\frac u{2w})^2 = \left(\frac u{2w}\right)^2$$ and transform it back by $M$: $$(x,y)\ \text{ in the red ellipse} \iff M^{-1}\begin{bmatrix} x\\y\\1 \end{bmatrix}\ \text{in the circle} $$