Prove $X^2+Y^2-1$ is irreducible using geometrical tools. I'm trying to understand what is meant in this paragraph:
 of "Conics and Cubics. A Concrete Introduction to Algebraic Curves (by Robert Byx)": 

He wants to prove that the polynomial $X^2+Y^2-1$ is irreducible using geometrical  tools. I have the following doubts:


*

*What he means by Since every line has at least three points on it ?

*Why the fact that $X^2+Y^2-1=F(X,Y)\cdot G(X,Y)$, where F, G are lines implies that the circle $X^2+Y^2-1=0$ is made up of two lines?
I would really grateful if anyone could help me.

*

*A line in $k^2$ has the form $aX+bY+c=0$ where $a$ and $b$ cannot both be zero.  We may assume without loss of generality that $a\ne 0$, so that we have $X=-a^{-1}bY-a^{-1}c$.  It follows that for any $\alpha\in k$, the point
$$(-a^{-1}b\alpha-a^{-1}c,\alpha)$$
lies on the line.  It follows that the number of points on any line is in bijective correspondence with the field $k$.  So in order for a line to have at least three points, we should assume that $k$ is not the field of $2$ elements, which is not a very strict assumption.

*If $X^2+Y^2-1=F(X,Y)\cdot G(X,Y)=0$, then notice by the zero product property that any point on the circle must satisfy at least one of the linear polynomials $F(X,Y)$ and $G(X,Y)$.  It is also true that any point satisfying the polynomials $F$ and $G$ will lie on the circle.  It follows that the circle must be the union of two lines (a contradiction).

