Let's consider the sphere centered at point $(0,0,-1)$ which is defined by the equation $x^2+y^2+(z+1)^2=4$. We are interested in a parametric expression of the intersection of that sphere with the plane $x=z$.
I have already found many questions here on MSE on this or related problems but they rely on a trick where you do some manipulations that finally lead to a nice quadratic form. In my case I haven't found such a trick.
Manipulating the equations I get: $$ 2x^2+2x+1+y^2=4. $$ It resembles an ellipsis but I don't know how to get rid of the $2x$ term. What I have found out by playing around with the parametrization of an ellipsis is that $\varphi:[0,2\pi]\to\mathbb{R}^3$, where $$ \varphi(t)=\begin{cases}\varphi_1(t)=\frac{2}{\sqrt{2}}\cos(t)\\\varphi_2(t)=2\sin(t)\\\varphi_3(t)=-1+\frac{2}{\sqrt{2}}\cos(t),\end{cases} $$ satisfies the equation $2x^2+2x+1+y^2=4.$ But how do I know that the converse is true, i.e. that a solution of $2x^2+2x+1+y^2=4$ can be always represented by $\varphi$?