I am confused because I know an ellipse has two foci at $(c,0)$ and $(-c,0)$. That would mean $c=0$, so $a^2=b^2$, assuming the standard equation of an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. This would mean $x^2+y^2=a^2$ which is the equation of a circle centered at the origin with radius $a$.
I have been told the answer is $r(\theta) = \frac{ed}{1+ecos(\theta)}$
This requires you to find the polar coordinates $r(\theta)$ drawn from one of the foci.
This means the cartesian to polar conversion is
$$ \pmatrix{x \\ y} = \pmatrix{ c + r \cos \theta \\ r \sin \theta} $$
with $c = \sqrt{a^2-b^2}$
Now use the above in the equation of the ellipse $$ \left( \frac{x}{a} \right)^2 + \left( \frac{y}{b} \right)^2 = 1$$ to get
$$ r(\theta) = \frac{b^2 (c \cos \theta -a)}{(a^2-b^2) \cos^2 \theta-a^2} $$
But also use the eccentricity $\epsilon = \sqrt{1 - \left( \tfrac{b}{a} \right)^2}$, as well as $c = \epsilon\; a$ to simplify the above into
$$ r(\theta) = \frac{a (1-\epsilon^2)}{1+\epsilon \cos \theta} $$