I have a system of $2$ non-linear equations that has more than according to the variation of one parameter have $2$, $1$ or $0$ solutions. I would like to add a 3rd equation to the system so the extra variable give me as result the parameter to have only $1$ solution.
I give you an example:
$x^2 + y^2 - 0.8x + p = 0,$
$-x^2 + 2x - 2y - p^2 = 0$.
In the system above, $x$ and $y$ are the variable and $p$ is the parameter. What equation should I add to the system to find the value of the parameter $p$ that gives me only $1$ solution of the system?
I would take the second equation and solve this one for $y$ $$y=\frac{1}{2}(-x^2+2x-p^2)$$ and plug this one in the first equation: $$1/4\,{p}^{4}+1/2\,{p}^{2}{x}^{2}+1/4\,{x}^{4}-{p}^{2}x-{x}^{3}+2\,{x}^ {2}+p-4/5\,x =0$$ The degree is four and the solution is not so simple.