Parametric question of the curve $x^2 + y^2 + 2x - 4y = 0$? What is the parametric form of the curve above?  If I had to solve it, what I would say is that the first step is to complete the square. However, where would I go from there?
 A: $$x^2 + y^2 + 2x - 4y = 0$$
$$x^2  + 2x +  y^2- 4y = 0$$
$$x^2  + 2x +  1 -1 +y^2- 4y +4 -4= 0$$
$$(x+1)^2 -1 + (y-2)^2-4= 0$$
$$(x+1)^2  + (y-2)^2= 5$$
For a circle with centre at (h,k) and radius r, the a parametric form is $x = h + r\cos(\theta)$ and $y = k + r\sin(\theta)$.
A: You can give a rational parametrization using stereographical projection, as follows:
The point $(0,0)$ is on the curve. Since the equation is quadratic, for each slope $t$, the line $y=tx$ must intersect the curve at a second point. If you plug $y=tx$ into the equation (and you assume $x\neq 0$), then you should get that the coordinates of the second point of intersection are
$$x=\frac{2(2t-1)}{1+t^2},\quad y=\frac{2t(2t-1)}{1+t^2}.$$
Are those all the points on the curve? No, but almost all. If $(x_0,y_0)\neq (0,0)$ is on the curve, then consider the line $L$ from $(0,0)$ to $(x_0,y_0)$:


*

*If the line $L$ is vertical, then $x_0=0$, and from the equation we have 
$$y_0^2-4y_0=0$$
Thus, $y_0=0$ (not allowed, by assumption), or $y_0=4$. Hence $(x_0,y_0)=(0,4)$.

*Otherwise, the line $L$ is not vertical, and $L$ is of the form $y=tx$ and $(x_0,y_0)$ is as given above:
$$(x_0,y_0)=\left(\frac{2(2t-1)}{1+t^2},\ \frac{2t(2t-1)}{1+t^2}\right).$$
Hence, the points on your curve are 
$$\{(0,4)\}\cup \left\{\left(\frac{2(2t-1)}{1+t^2},\ \frac{2t(2t-1)}{1+t^2}\right): t\in \mathbb{R}\right\},$$
and that's the parametrization we were looking for (notice $(0,0)$ is on the right hand side, with $t=1/2$).
