Consider the two paraboloids $z=2-x^2-y^2$ and $z=x^2-2x+y^2-4y$. What I am trying to do is parametrise the intersection of the paraboloids. My original thoughts were that since both paraboloids are defined by $z$, I could go ahead and let $z=z$
Such that
$$2-x^2-y^2=x^2-2x+y^2-4y$$
By rearranging this mess by completing some squares and doing some multiplication and division I arrive at some circular equation
$$\left(x-\frac{1}{2}\right)+(y-1)^2=\frac{9}{4}$$
Which describes a circle with center at $\left(\frac{1}{2},1\right)$ with a radius $\frac{3}{2}$. So I parametrised that as
$$\textbf{r}(t)=\left(\frac{1}{2}+\frac{3}{2}\cos t,1+\frac{3}{2}\sin t\right)$$
But when I have a look at the figures in geogebra it seems very much like the intersection between the two paraboloids is an ellipse with a projection (which I am also looking for) in the xy-plane that looks like this parametrised circle. I just can't seem to find where it is that I am missing the mark, so any pointers would be greatly appreciated!
The intersection is a curve in the space. You have parametrized its projection on the XY-plane. So, your parametrization should include the height (the z-coordinate in terms of t). Hence, $$r(t)=\left(\frac{1}{2}+\frac{3}{2}\cos(t),1+\frac{3}{2}\sin(t),2-(\frac{1}{2}+\frac{3}{2}\cos(t))^2-(1+\frac{3}{2}\sin(t))^2\right).$$