Suppose that we use our compass to draw a circle whose center is a point that has coordinates from a field $K$ with radius from $K$ then we get a circle with equation $x^2+y^2+ax+by+c=0$. Similarly, if we use our straightedge to draw a line between two points with coordinates from $K$ we get a line with equation $ax+by+c=0$ .
Prove that, if there is a point of intersection, then its coordinates must be in a field of the form $K(\sqrt{d})$ where $d$ is some element of $K$.
Could you please help me with that?
Hint:
The intersection(s), if they exists, are solutions of the system $$\begin {cases} x^2+y^2+ax+by+c=0\\ px+qy+n=0 \end{cases} $$ (I used $p,q,m$ for the straight line to avoid confusion).
So they are the solution of a second degree equation $P^{(2)}(x)=Ax^2+Bx+C=0$ and have the form $ x= -\frac{B}{2A}\pm \frac{\sqrt{D}}{2A}$.