<s>[INST] Why combining two quadratic equations of a circle and a parabola creates extra solutions for $x$ We have a parabola and a circle with the following equations and their graph placed at the end of my question.
Parabola: $y^2 = 4x -4$
Circle: $(x-2)^2 + y^2 = 9$
My goal was to calculate their intersection points so I substituted $y^2$ from the parabola equation into the circle equation and I got
$(x-2)^2 + (4x-4)=9 \implies x^2 - 4x + 4 + (4x - 4) = 9 \implies x^2 = 9 \implies x = \pm3$
$x=3$ is the only correct solution but why is $x=-3$ produced as an extra invalid solution?
What is the exact mathematical explanation behind this? Why substituting one equation into the other has produced extra answers?

update
When I calculate $x$ from the parabola equation and substitute it in the circle equation, I don't get any extra answers for $y$:
$y^2=4x-4 \implies y^2 +4 = 4x \implies x = \frac{y^2}{4} + 1$
$(x-2)^2 +(4x-4)=9 \implies ((\frac{y^2}{4} + 1) - 2)^2 + (4x - 4)=9 \implies y^4 +8y - 128 = 0 \implies y^2=8,-16$
$y^2 = -16$ cannot be true so $y^2 = 8 \implies y=\pm 2\sqrt{2}$ and these are correct answers for $y$.
2nd update
I made a mistake in the calculation in the previous update although the final solutions where correct. I write the correct calculation:
$(x-2)^2 +y^2=9 \implies 
((\frac{y^2}{4} + 1) - 2)^2 + y^2=9 \implies 
(\frac{y^2}{4} - 1)^2 + y^2=9 \implies 
(\frac{y^4}{16} - \frac{y^2}{2} + 1) + y^2=9 \implies 
\frac{y^4}{16} + \frac{y^2}{2} + 1=9 \implies 
(\frac{y^2}{4} +  1)^2=9 \implies 
(\frac{y^2}{4} +  1)=\pm3 \implies
\frac{y^2}{4} =2,-4 \implies
y^2=8,-16$
 [/INST] I suppose that we are working on the real number system and not consider complex numbers but I think my reasoning works in complex numbers too. We want to solve a system of equations containing two equations. I move all terms of each equation to one side and name them $A(x,y)$ and $B(x,y)$
$\begin{cases}
      A(x,y)=y^2 -4x +4=0\\
      B(x,y)=y^2 +(x-2)^2-9=0\\
    \end{cases}$
We want to find tuples like $(x', y')$ such that $A(x',y')=0$ and $B(x',y')=0$ simultaneously.
If we calculate $y^2$ from $A(x,y)=0$ and substitute it into $B(x,y)$ we arrive at a third equation $C(x)=0$
$C(x) = 4x-4 +(x-2)^2-9 = 0 \implies C(x) = x^2 -9=0$
And when we solve $C(x) = 0$ we get $x^2=9 \implies x=\pm 3$ but any point $(-3,y)$ with $y\in\mathbb{R}$ does not satisfy $\begin{cases}
      A(x,y)=0\\
      B(x,y)=0\\     
    \end{cases}$
and the solving procedure has produced extraneous solutions. The reason is this the line of reasoning is not reversible.
$\begin{cases}
      A(x,y)=0\\
      B(x,y)=0\\     
    \end{cases} \overset{1}{\implies} C(x)=0 \overset{2}{\iff} x=\pm 3$
The $\overset{1}{\implies}$ is not reversible. In this case when there exists $(x_0,y_0)$ as a solution to the system we will have:
$\begin{cases}
      A(x_0,y_0)=0\\
      B(x_0,y_0)=0\\     
    \end{cases} \overset{3}\implies C(x_0)=0 \overset{4}\iff x_0=\pm 3$
but if there exits $(x_1,y_1)$ as a solution to $C(x) = 0$ we will have
$C(x_1)=0 \;\not\!\!\!\implies \begin{cases}
      A(x_1,y_1)=0\\
      B(x_1,y_1)=0\\     
    \end{cases}$
When we combine the equations we loose information and cannot retrive the system $\begin{cases}
      A(x_1,y_1)=0\\
      B(x_1,y_1)=0\\     
    \end{cases}$ from $C(x)=0$
so every solution to the system is a solution to $C(x)=0$ but we cannot say every solution to $C(x)=0$ must be a solution to the system and extraneous solutions might have been produced.
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