 Solution

Consider an isolated intersection of the two curves. For two distinct lines, solving their linear equations by elimination gives coordinates in $F$ whenever there is an intersection.

For a line and a circle, solve the line for one coordinate with a nonzero coefficient and substitute into the circle equation. The resulting quadratic has coefficients in $F$. A real root lies in $F$ if its discriminant is a square in $F$, and otherwise in $F(\sqrt{D})$ for its positive nonsquare discriminant $D\in F$. The other coordinate is an $F$-linear expression in that root. Thus both coordinates lie in the same real quadratic extension. A tangent intersection has discriminant zero and already lies in $F$.

Write two circle equations in normalized form $x^2+y^2+u_{i}x+v_{i}y+w_i=0$. Subtracting eliminates the quadratic terms. If the circles are distinct and meet, this gives a nontrivial line, reducing the problem to the preceding case. Concentric distinct circles do not intersect. ThereforeThe isolated-intersection qualification is implicit in the construction operation. If coincident curves were allowed and every point on their common curve counted as an intersection point, the printed assertion would be false: a rational line contains points with transcendental coordinates.