<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] A standard Cartesian graph deals only with real numbers "plotted against" real numbers, so in looking at the intersections of two curves in such a graph, one may not be "getting the whole story."   Two distinct quadratic curves will have four intersections; since one of these (the circle) is bounded, all of the intersection points will be at finite distances from the origin.
If we use the curve equations with complex numbers $ \ x \ = \ a + bi \ $ and $ \ y \ = \ c + di \ \ , $ (with $ \ a \ , \ b \ , \ c \ , \ d \ $ real), they  become
$$ (x-2)^2 \ + \ y^2 \ \ = \ \ 9 \ \ \rightarrow \ \ ( \ [a - 2] + bi \ )^2 \ + \ ( \ c \ + \ di \ )^2 \ \ = \ \ 9 $$
$$ \rightarrow \ \ [ \ (a^2 - 4a - b^2 + 4) \ + \ (2a  - 4 )·b·i \ ]   \ + \ [ \ (c^2 - d^2) \ + \ 2cd·i \ ] \ \ = \ \ 9 \ + \ 0·i  \ \ ; $$
$$ y^2 \ \ = \ \ 4x \ - \ 4 \ \ \rightarrow \ \ (c + di)^2 \ \ = \ \ 4·(a + bi) \ - \ 4 \ \  = \ \ (4a - 4) \ + \ 4bi \ \ . $$
The calculation without the imaginary parts ($ \ b \ = \ d \ = \ 0 \ $) is the one you made (the one we learn to make in "standard" analytic geometry), giving $ \ \ a^2 \ = \ 9 \ \ , \ \ c^2 \ = \ 4·a - 4   \ \ . $   We take $ \ a \ = \ +3 \ \ , $ $ c^2 \ = \ 4·(+3) - 4 \ = \ 8 \ \ , $ and "discard"   $ \ a \ = \ -3 \ $ , since it leads to $ \ c^2 \ = \ 4·(+3) - 4 \ = \ -16 \ \ $ , which is not permissible since $ \ c \ $ is taken to be a real number.  Your graph then represents real part $ \ c \ $ for $ \ y \ $ as a function of real part $ \ a \ \   $ for $ \ x \ \ . $
But a "graph" using complex numbers needs to be made in $ \ \mathbb{C}^2 \ \ , $ which can be interpreted in $ \ \mathbb{R}^4 \ \ , $ making it challenging to visualize, as the "curves" are then treated as four-dimensional.  The  intersections we "rejected" occur in those extended parts of the curves.  If we return to the curve equations and work with the two-dimensional "slice" that "graphs" the imaginary part $ \ d \ $ of $ \ y \ $ against the real part $ \ a \ $ of $ \ x \ \ $ (so $ \ b \ = \ c = \ 0 \ $ ) , we obtain
$$ ( \ [a - 2] + 0·i \ )^2 \ + \ ( \  0 + di \ )^2 \ \ = \ \ 9 \ \ \rightarrow \ \ (a - 2)^2 \ -  \ d^2 \ \ = \ \ 9 \ \ ; $$
$$ (0 + di)^2 \ \ = \ \ 4·(a + 0·i) \ - \ 4 \ \  \rightarrow \ \ -d^2 \ \ = \ 4a \ - \ 4  \ \ . $$
This produces $ \ a^2 \ = \ 9 \ \ , \ \ -d^2 \ = \ 4a - 4 \ \ , $ but using $ \ a \ = \ +3 \ $ will give us $ \ -d^2 \ = \ 8 \ \ , $ which we now "reject" because $ \ d \ $ is a real number; here, we take $ \ a \ = \ -3 \ \Rightarrow \ -d^2 \ = \ -16 \ \Rightarrow \ d \ = \ \pm 4 \ \ . $  As we see on the graph of $ \ d \ $ versus $ \ a \ $ below, we have found that a hyperbolic part of $ \ (x-2)^2   +   y^2 \   =   \ 9 \ $ intersects a parabolic part of $ \ y^2 \ \ = \ \ 4x \ - \ 4 \ $ at the points $ \ (-3 \ , \ \pm   4i) \ \ , $ as mentioned by Narasimham.

We have accounted then for the four intersections of the two curves, two of which are "invisible" on a "standard" graph in $ \ \mathbb{R}^2 \ \ . $  No other intersections appear on other "slices":  for instance, the imaginary part $ \ d \ $ versus imaginary part $ \ b \ $ equations are
$$ ( \ [0 - 2] + b·i \ )^2 \ + \ ( \  0 + di \ )^2 \ \ = \ \ 9 \ \ \rightarrow \ \ 4 \ - \ b^2 \ -  \ d^2 \ \ = \ \ 9 \ \ \rightarrow \ \ b^2 \ + \ d^2 \ \ = \ \ -5 \ \ ; $$
$$ (0 + di)^2 \ \ = \ \ 4·(0 + b·i) \ - \ 4 \ \  \rightarrow \ \ -d^2 \ \ = \ 4bi \ - \ 4  \ \ , $$
which do not produce real solutions for $ \ b \ $ and $ \ d \ \ . $
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