Conceptual question about constructible points of circle-line intersections in the context of field extensions

Question

In Chapter 30 of Pinter's "A Book of Abstract Algebra", the reader is introduced to the idea of constructible points (using a ruler and compass) and how they can be viewed in relation to field extensions and vector spaces. I have a conceptual question about the field extensions that are associated with constructible points derived from the intersection of a line segment with a circle

Besides the pictures I will provide below, the only necessary information is as follows:

1. Let $P$ be a constructible point within the the plane $\mathbb Q \times \mathbb Q$.

2. Suppose $P$ has coordinates $(a,b)$. We associate with $P$ the field $\mathbb Q (a,b)$, which is obtained by adjoining the coordinates of $P$ to $\mathbb Q$.

3. Let $K_1$ be the resulting field of $\mathbb Q (a,b)$...that is to say, $K_1 = \mathbb Q (a,b)$.

4. For each additional constructible point $P_i = (a_i,b_i)$ that is associated to a pre-existing $K_{i-1}$, let $K_i = K_{i-1}(a_i,b_i)$

With those in mind, here is the lemma and a portion of the accompanying proof which inspired the question:

I have outlined in $\color{#c00}{red}$ the portion of the proof that I want to focus on:

Specifically, if Point $S$ is defined by the coordinates $(x_i,y_i)$ and Point $T$ is defined by the coordinates $(x_j,y_j)$, in general, are the field extensions $K_{i-1}(x_i,y_i)$ and $K_{i-1}(x_j, y_j)$ equivalent? That is to say, there are two $x$ values that will serve as the roots of the quadratic polynomial that solve for points of intersection between the line and circle: $x_i$ and $x_j$. With their accompanying $y$ values ($y_i$ and $y_j$, respectively), are these two extensions, formed by associating different but related constructible points $S$ and $T$ to the same pre-existing field $K_{i-1}$ generally equivalent?

My intuition says no, besides two special cases:

Case 1: The line segment is coincident with a diameter line

Case 2: The line segment intersects the circle such that the interior angle is $90 \deg$

Any commentary is greatly appreciated!

Answer 1

Yes, you get the same extensions from both points. The reason is the structure of the solutions of the quadratic equation. Say for $x_i$ you get two solutions of the form $$ x_i=\frac{-b\pm\sqrt{\Delta}}{2a} $$ which both belong to the same extension $\mathbb{Q}\cup\{\sqrt{\Delta\}}$ (assuming there are real solutions of course). The same is true about $y_i$. Therefore, the extensions are the same.