Number of common zeros of two quadratic polynomials in ${\Bbb C}[t,x]$ The following theorem is in Artin's Algebra(2nd edition):

Theorem 11.9.10 Two nonzero polynomials $f(t,x)$ and $g(t,x)$ in two variables have only finitely many common zeros in ${\Bbb C}^2$, unless they have a common nonconstant factor in ${\Bbb C}[t,x]$.

A comment after this theorem says that "It is harder to prove the Bezout bound than the finiteness. We won't need that bound, so we won't prove it". 
Then there is an exercise in the 1st edition:

Prove that two quadratic polynomials $f, g$ in two variables have at most four common zeros, unless they have a nonconstant factor in common.

Then in the 2nd edition, this exercise has been changed to the following one:

Let $C_1$ and $C_2$ be the zeros of quadratic polynomials $f_1$ and $f_2$ respectively that don't have a common linear factor.
  (a) Let $p$ and $q$ be distinct points of intersection of $C_1$ and $C_2$, and let $L$ be the (complex) line through $p$ and $q$. Prove that there are constants $c_1$ and $c_2$, not both zero, so that $g=c_1f_1+c_2f_2$ vanishes identically on $L$. Prove also that $g$ is the product of linear polynomials.
  (b)Prove that $C_1$ and $C_2$ have at most $4$ points in common. 

I think (a) is supposed to be a hint to (b). But I don't see how I can make the implication. 
I guess I have to show that there cannot be too many $(c_1,c_2)$ as in (a), while such pairs might correspond to zeros of a polynomial of small degree.

Here is my question:
How can I prove (b) using (a)?
 A: OK, the discussion in the comments is getting a little confusing, so let me write a full answer instead. I'll try to spell everything out in detail.
We are assuming statement (a) and trying to deduce (b). Of course, we can assume that $C_1 \cap C_2$ contains at least 4 points.
Claim 1: No 3 points of $C_1 \cap C_2$ lie on a line. 
Proof of claim: A nonzero quadratic polynomial in one variable has at most 2 roots. If 3 points $p,q,r$ of $C_1 \cap C_2$ lay on a line $L$, then both $f_1$ and $f_2$ would be identically zero on $L$. But then they would have a linear factor in common, contrary to hypothesis. $\ \square$
Let $p,q$ be two points of $C_1 \cap C_2$, and let $L_1 =\overline{pq}$,
By Part (a) we can find numbers $c_1, \ c_2$ such that $g_1=c_1f_1+c_2f_2$  vanishes on $L_1$ and factors into linear polynomials, say $g_1=l_1 \lambda_1$, where $l_1$ vanishes on $L_1$.
Now choose $r$ to be a point of $C_1 \cap C_2$ such that neither $l_1$ or $\lambda_1$ vanishes on $\overline{pr}$. (If $\lambda_1$ doesn't vanish at $p$, any point other than $p$ or $q$ will do; if it does, then by Claim 1 $V(l_1) \cup V(\lambda_1)$ contains at most 3 points of $C_1 \cap C_2$, so we can take $r$ to be any other point of $C_1 \cap C_2$.) 
Let $L_2=\overline{pr}$; then applying (a) again we can find numbers $d_1, \ d_2$ such that $g_2=d_1f_1+d_2f_2$  vanishes on $L_2$ and factors into linear polynomials.
Claim 2: The matrix
$$ \begin{pmatrix}
c_1 & c_2 \\
d_1 & d_2
\end{pmatrix}$$
is invertible. 
Proof of claim: From Part (a) both rows of the matrix are nonzero. Suppose there exists $\lambda$ such that $(\lambda d_1, \lambda d_2)=(c_1,c_2)$. Then $g_1=\lambda g_2$. But then $g_1$ must vanish on $L_2$, contrary to the choice of $L_2$. $\ \square$
So now we can write 
$$\begin{align*}
\begin{pmatrix}
g_1 \\ g_2 
\end{pmatrix} &=
M \begin{pmatrix}
f_1 \\ f_2 
\end{pmatrix}
\end{align*}$$
for an invertible matrix $M$. This shows that $V(f_1,f_2)=V(g_1,g_2)$; moreover, it also shows that $g_1$ and $g_2$ cannot share a linear factor, since then $f_1$ and $f_2$ would also, contrary to hypothesis.
So we can write $g_1=l_1 \lambda_1$, $g_2= l_2\lambda_2$ where the $l_i$ and $\lambda_i$ are linear polynomials, and no linear term appears in both factorisations.
So now 
$$V(f_1,f_2) = V(g_1,g_2) = \bigcup_{i,j=1,2} V(l_i) \cap V(\lambda_j).$$
Each of the intersections consists of at most 1 point, so the union consists of at most 4 points. 
