<s>[INST] What causes the extraneous root when intersecting parabola $y^2=4ax$ with circle $x^2+y^2=9a^2/4$? If I solve the parabola: $y^2=4ax$ and the circle: $x^2+y^2=\frac{9a^2}{4}$ I get a quadratic in $x$; ie: 
$$4x^2+16ax-9a^2=0$$ 
which has roots $x=\frac{a}{2},-\frac{9a}{2}$. But if we see the graphs of these two, both the intersections occur at same $x$-coordinate so why does this extra root $x=-\frac{9a}{2}$ come?
So my question is that if the circle was unknown say $x^2+y^2=b^2$ then I would have put the discriminant of the quadratic in $x$ as $0$ but that would be wrong so how do I judge that an extraneous root will be present and discriminant won't be zero?
 [/INST] The full solution is this $$\begin{cases}x^2+y^2=\frac94a^2\\ y^2=4ax\end{cases}\iff \begin{cases}x^2+4ax-\frac94a^2=0\\ y^2=4ax\end{cases}\iff\begin{cases}x=\frac a2\\ y^2=2a^2\end{cases}\lor\begin{cases}x=-\frac 92a\\ y^2=-18a^2\end{cases}$$
Now the system must be "discussed" in terms of the parameter, in the sense that it is equivalent to the following: $$\begin{cases}x=\frac a2\\ y^2-2a^2=0\end{cases}\lor\begin{cases}x=-\frac 92a\\ y^2+18a^2=0\end{cases}\iff\begin{cases}x=\frac a2\\ y=\sqrt2a\end{cases}\lor\begin{cases}x=\frac a2\\ y=-\sqrt2a\end{cases}\lor\begin{cases}x=0\\ y=0\\ a= 0\end{cases}$$
And you decide whether or not to accept the degenerate case.
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