Since the circumference is the place of the points distant $R > 0$ from a point $\left(x_c, \, y_c\right)$, its cartesian equation is:
$$
\left(x - x_c\right)^2 + \left(y - y_c\right)^2 = R^2
$$
whose rational parameterization is $\left(x, \, y\right) = \left(x_c + R\,\frac{1-z^2}{1+z^2}, \, y_c + R\,\frac{2\,z}{1+z^2}\right)$, with $z \in \mathbb{R}$.
At this point, an ellipse assigned in canonical form:
$$\frac{\left(x - x_c'\right)^2}{a^2} + \frac{\left(y - y_c'\right)^2}{b^2} = 1$$
with $a, \, b > 0$, substituting for $x, \, y$ the respective parametric equations of the circumference is obtained:
$$
\frac{\left(x_c + R\,\frac{1-z^2}{1+z^2} - x_c'\right)^2}{a^2} +
\frac{\left(y_c + R\,\frac{2\,z}{1+z^2} - y_c'\right)^2}{b^2}
= 1
$$
ie:
$$
\small
\left(x_c - x_c' + R\,\frac{1-z^2}{1+z^2}\right)^2\,b^2\left(1+z^2\right)^2 +
\left(y_c - y_c' + R\,\frac{2\,z}{1+z^2}\right)^2\,a^2\left(1+z^2\right)^2
= a^2\,b^2\left(1+z^2\right)^2
$$
from which:
$$
\small
b^2\left(x_c - x_c'\right)^2\left(1 + z^2\right)^2 +
2\,b^2\,R\left(x_c - x_c'\right)\left(1 - z^4\right) +
b^2\,R^2\left(1 - z^2\right)^2 + \\
\small
a^2\left(y_c - y_c'\right)^2\left(1 + z^2\right)^2 +
2\,a^2\,R\left(y_c - y_c'\right)\left(2\,z\right)\left(1 + z^2\right) +
a^2\,R^2\left(2\,z\right)^2
= a^2\,b^2\left(1+z^2\right)^2
$$
which simplified is equivalent to the equation:
$$
a_4\,z^4 + a_3\,z^3 + a_2\,z^2 + a_1\,z + a_0 = 0
$$
with:
$$
\begin{aligned}
& a_4 = a^2\left(\left(y_c - y_c'\right)^2 - b^2\right) + b^2\left(\left(x_c - x_c'\right) - R\right)^2 \\
& a_3 = 4\,a^2\,R\left(y_c - y_c'\right) \\
& a_2 = 2\left(a^2\left(\left(y_c - y_c'\right)^2 - b^2 + 2\,R^2\right) + b^2\left(\left(x_c - x_c'\right)^2 - R^2\right)\right) \\
& a_1 = 4\,a^2\,R\left(y_c - y_c'\right) \\
& a_0 = a^2\left(\left(y_c - y_c'\right)^2 - b^2\right) + b^2\left(\left(x_c - x_c'\right) + R\right)^2
\end{aligned}
$$
All that remains is to solve this equation and replace the real solutions in the parametric equations of the circumference to obtain the coordinates of the desired points of intersection.
Furthermore, in the event of:
$$
\Delta := \left(\left(x_c - x_c'\right) - (R - a)\right)\left((R + a) - \left(x_c - x_c'\right)\right) \ge 0
$$
and:
$$
a\left(y_c - y_c'\right) - b\,\sqrt{\Delta} = 0
\; \; \; \vee \; \; \;
a\left(y_c - y_c'\right) + b\,\sqrt{\Delta} = 0
$$
then the circumference and the ellipse intersect in the singular point $\left(x_c - R, \; y_c\right)$.
This algorithm can be implemented relatively easily in Microsoft Excel as follows:
referring, for example, to what is reported in the following excellent books:
- Handbook of Mathematical Functions with Formulas, Graphs and Mathematical Tables - Abramowitz, Stegun - USA
- Handbook of Mathematical Functions - Frank W. J. Olver - NIST
Best Answer
As noted in the answer of @rfabbri ( +1), the perspective projection of a circle is not always an ellipse, but it is in general a conic section (maybe degenerate).
If you want a matrix that transform the circle in an ellipse by projection, than note that teh general equation of a conic: $$ Ax^2+Bxy+Cy^2+Dx+Ey+F=0$ $$
can be written in the form (see here): $$ \begin{pmatrix} x&y&1 \end{pmatrix} \begin{pmatrix} A&\frac{B}{2}&\frac{D}{2}\\ \frac{B}{2}&C&\frac{E}{2}\\ \frac{D}{2}&\frac{E}{2}&F\\ \end{pmatrix} \begin{pmatrix} x\\y\\1 \end{pmatrix}=0 $$
so, for $A=C=1$ and $B,D,E=0$ and $F=-r^2$ we have the circle center at the origin and radius $r$, that becomes an ellipse if the matrix is such that $B^2-4AC <0$ and the conic is not degenerate, i.e. the determinat of the matrix is not null.