[proofplan] Set $K := \mathbb{Q}(S)$, the subfield of $\mathbb{R}$ generated over $\mathbb{Q}$ by all coordinates of points in $S$. A point $P$ is 1-step constructible from $S$ if $P$ is an intersection point of two objects (lines and circles) determined by points in $S$. Lines through pairs of points in $S$ have equations with coefficients in $K$, and circles centred at points of $S$ with radii equal to distances between points of $S$ have equations with coefficients in $K$. We analyse the three cases — line-line, line-circle, circle-circle — separately. In each case, finding the coordinates of $P$ reduces to solving equations over $K$: the line-line case yields a linear system (degree 1), the line-circle case yields at most a quadratic equation (degree 1 or 2), and the circle-circle case reduces to the line-circle case by subtracting one circle equation from the other. [/proofplan]

[step:Set up notation and express lines and circles as equations with coefficients in $K$]Write $K := \mathbb{Q}(S)$, where $\mathbb{Q}(S)$ denotes the subfield of $\mathbb{R}$ generated over $\mathbb{Q}$ by the coordinates of all points in $S$. Since $S \subset \mathbb{R}^2$ is finite, $K$ is a subfield of $\mathbb{R}$. A point $P = (p_1, p_2) \in \mathbb{R}^2$ is 1-step constructible from $S$ if $P$ arises as an intersection point of two distinct objects from the following list: - **Lines** through pairs of distinct points in $S$, - **Circles** centred at a point of $S$ with radius equal to the distance between two points of $S$. We express these objects as equations in unknowns $(x, y)$ with coefficients in $K$. **Lines.** Let $A = (a_1, a_2)$ and $B = (b_1, b_2)$ be distinct points of $S$. The line through $A$ and $B$ has equation \begin{align*} (b_2 - a_2)(x - a_1) - (b_1 - a_1)(y - a_2) = 0, \end{align*} which expands to a linear equation $\alpha x + \beta y + \gamma = 0$ with $\alpha, \beta, \gamma \in K$ (since $a_1, a_2, b_1, b_2 \in K$). The coefficients $\alpha = b_2 - a_2$ and $\beta = -(b_1 - a_1)$ are not both zero because $A \neq B$. **Circles.** Let $C = (c_1, c_2) \in S$ be the centre and let $A = (a_1, a_2), B = (b_1, b_2) \in S$ determine the radius $r = |AB|$. The circle has equation \begin{align*} (x - c_1)^2 + (y - c_2)^2 = (a_1 - b_1)^2 + (a_2 - b_2)^2. \end{align*} Since $a_1, a_2, b_1, b_2, c_1, c_2 \in K$, all coefficients of this equation lie in $K$. Expanding, we may write this as \begin{align*} x^2 + y^2 + \delta x + \varepsilon y + \zeta = 0, \end{align*} with $\delta, \varepsilon, \zeta \in K$. The coordinates $(p_1, p_2)$ of $P$ satisfy a system of two such equations. We analyse the three possible pairings.[/step]

[guided]The first task is to translate the geometric definition of 1-step constructibility into algebra. The key observation is that every geometric object determined by points of $S$ — whether a line or a circle — is described by a polynomial equation whose coefficients lie in $K = \mathbb{Q}(S)$. Why is this? The coordinates of all points in $S$ lie in $K$ by definition, and the coefficients of the line and circle equations are constructed from these coordinates using only addition, subtraction, and multiplication (which are the field operations in $K$). For lines, the equation $(b_2 - a_2)(x - a_1) - (b_1 - a_1)(y - a_2) = 0$ involves only differences and products of coordinates of $A$ and $B$. For circles, the centre coordinates and the squared radius $(a_1 - b_1)^2 + (a_2 - b_2)^2$ are similarly polynomial expressions in elements of $K$. This algebraic reformulation is what allows us to use field extension theory: the question "what field do the coordinates of $P$ generate?" reduces to "what field do the solutions of polynomial equations over $K$ lie in?"[/guided]

[step:Case 1: Intersection of two lines gives degree 1]Suppose $P$ is the intersection of two distinct lines, each passing through a pair of points of $S$. The coordinates $(p_1, p_2)$ of $P$ satisfy the system \begin{align*} \alpha_1 x + \beta_1 y + \gamma_1 &= 0, \\ \alpha_2 x + \beta_2 y + \gamma_2 &= 0, \end{align*} where $\alpha_i, \beta_i, \gamma_i \in K$ for $i \in \{1, 2\}$, and the two lines are distinct. Since the two lines are distinct and intersect (at $P$), the coefficient matrix \begin{align*} M := \begin{pmatrix} \alpha_1 & \beta_1 \\ \alpha_2 & \beta_2 \end{pmatrix} \end{align*} has $\det M = \alpha_1 \beta_2 - \alpha_2 \beta_1 \neq 0$ (two distinct lines in $\mathbb{R}^2$ that intersect in a point are non-parallel, hence have linearly independent normal vectors). By Cramer's rule, \begin{align*} p_1 &= \frac{\beta_1 \gamma_2 - \beta_2 \gamma_1}{\alpha_1 \beta_2 - \alpha_2 \beta_1}, & p_2 &= \frac{\alpha_2 \gamma_1 - \alpha_1 \gamma_2}{\alpha_1 \beta_2 - \alpha_2 \beta_1}. \end{align*} Since all quantities on the right-hand side lie in $K$ and the denominator is nonzero, both $p_1$ and $p_2$ lie in $K$. Therefore $K(p_1, p_2) = K$, giving \begin{align*} [\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] = [K(p_1, p_2) : K] = 1. \end{align*}[/step]

[guided]The line-line case is the simplest: a system of two linear equations over a field $K$ with a unique solution produces a solution in $K$ itself. No field extension is needed. More precisely, the two lines are distinct and intersect, which means they are not parallel. Two lines $\alpha_1 x + \beta_1 y + \gamma_1 = 0$ and $\alpha_2 x + \beta_2 y + \gamma_2 = 0$ in $\mathbb{R}^2$ are parallel if and only if the normal vectors $(\alpha_1, \beta_1)$ and $(\alpha_2, \beta_2)$ are proportional, i.e., $\alpha_1 \beta_2 - \alpha_2 \beta_1 = 0$. Since the lines intersect in a unique point, $\det M \neq 0$, and Cramer's rule gives the unique solution as a ratio of elements of $K$. Note that $\mathbb{Q}(S \cup \{P\}) = K(p_1, p_2)$: the field $\mathbb{Q}(S \cup \{P\})$ is generated over $\mathbb{Q}$ by the coordinates of all points in $S$ together with $p_1$ and $p_2$, which equals the field generated over $K = \mathbb{Q}(S)$ by $p_1$ and $p_2$. Since $p_1, p_2 \in K$, the extension is trivial: $[K(p_1, p_2) : K] = 1$.[/guided]

[step:Case 2: Intersection of a line and a circle gives degree 1 or 2]Suppose $P$ is an intersection of a line and a circle, both determined by points of $S$. The coordinates $(p_1, p_2)$ of $P$ satisfy \begin{align*} \alpha x + \beta y + \gamma &= 0, \\ x^2 + y^2 + \delta x + \varepsilon y + \zeta &= 0, \end{align*} with $\alpha, \beta, \gamma, \delta, \varepsilon, \zeta \in K$ and $(\alpha, \beta) \neq (0, 0)$. Without loss of generality, assume $\beta \neq 0$ (the case $\alpha \neq 0$ is handled identically by interchanging the roles of $x$ and $y$). Solve the linear equation for $y$: \begin{align*} y = -\frac{\alpha x + \gamma}{\beta}. \end{align*} Since $\alpha, \beta, \gamma \in K$ and $\beta \neq 0$, this expresses $y$ as a degree-1 polynomial in $x$ with coefficients in $K$. Substituting into the circle equation: \begin{align*} x^2 + \left(-\frac{\alpha x + \gamma}{\beta}\right)^2 + \delta x + \varepsilon \left(-\frac{\alpha x + \gamma}{\beta}\right) + \zeta = 0. \end{align*} Multiplying through by $\beta^2 \neq 0$ and expanding: \begin{align*} \beta^2 x^2 + (\alpha x + \gamma)^2 + \delta \beta^2 x - \varepsilon \beta (\alpha x + \gamma) + \zeta \beta^2 &= 0, \\ (\alpha^2 + \beta^2) x^2 + (2\alpha\gamma + \delta\beta^2 - \varepsilon\alpha\beta) x + (\gamma^2 - \varepsilon\beta\gamma + \zeta\beta^2) &= 0. \end{align*} Define the coefficients \begin{align*} A &:= \alpha^2 + \beta^2, \\ B &:= 2\alpha\gamma + \delta\beta^2 - \varepsilon\alpha\beta, \\ C &:= \gamma^2 - \varepsilon\beta\gamma + \zeta\beta^2. \end{align*} All three lie in $K$. Since $(\alpha, \beta) \neq (0,0)$, we have $A = \alpha^2 + \beta^2 > 0$ (in $\mathbb{R}$), so $A \neq 0$ and the equation $Ax^2 + Bx + C = 0$ is a genuine quadratic over $K$. The quadratic formula gives \begin{align*} p_1 = \frac{-B \pm \sqrt{B^2 - 4AC}}{2A}, \end{align*} where $d := B^2 - 4AC \in K$ is the discriminant. We distinguish two sub-cases. **Sub-case 2a: $d$ is a perfect square in $K$** (i.e., $d = e^2$ for some $e \in K$). Then $p_1 = \frac{-B \pm e}{2A} \in K$, and $p_2 = -\frac{\alpha p_1 + \gamma}{\beta} \in K$. Therefore $K(p_1, p_2) = K$ and $[\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] = 1$. **Sub-case 2b: $d$ is not a perfect square in $K$.** Then $\sqrt{d} \notin K$, so $p_1 = \frac{-B + \sqrt{d}}{2A}$ or $p_1 = \frac{-B - \sqrt{d}}{2A}$. In either case, $K(p_1) = K(\sqrt{d})$, which is a degree-2 extension of $K$: the element $\sqrt{d}$ satisfies the polynomial $t^2 - d \in K[t]$, which is irreducible over $K$ because $d$ has no square root in $K$. By the [Structure of Simple Algebraic Extensions](/theorems/1251), $[K(\sqrt{d}) : K] = 2$. Since $p_2 = -(\alpha p_1 + \gamma)/\beta$ is a $K$-linear expression in $p_1$, we have $K(p_1, p_2) = K(p_1) = K(\sqrt{d})$. Therefore \begin{align*} [\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] = [K(p_1, p_2) : K] = 2. \end{align*} In both sub-cases, $[\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] \in \{1, 2\}$.[/step]

[guided]The line-circle case is the heart of the argument. The strategy is to use the linear equation to eliminate one variable, reducing the system to a single polynomial equation in one unknown. Assume $\beta \neq 0$ (if $\beta = 0$, then $\alpha \neq 0$ and we solve for $x$ instead; the argument is identical with the roles of $x$ and $y$ exchanged). Solving the line equation for $y$ gives $y = -(\alpha x + \gamma)/\beta$, a rational expression in $x$ with coefficients in $K$. Substituting into the circle equation eliminates $y$ and produces a polynomial equation in $x$ alone. We must verify that this equation has degree exactly 2 (not higher, and not degenerate to degree 0 or 1). Expanding, the coefficient of $x^2$ is $A = \alpha^2 + \beta^2$. Since $\alpha, \beta \in \mathbb{R}$ and $(\alpha, \beta) \neq (0, 0)$, we have $\alpha^2 + \beta^2 > 0$, so $A \neq 0$ and the equation is genuinely quadratic. Why does this matter? A quadratic equation $Ax^2 + Bx + C = 0$ over a field $K$ (with $A \neq 0$) has roots lying either in $K$ itself or in a quadratic extension $K(\sqrt{d})$, where $d = B^2 - 4AC$ is the discriminant. This is because the roots are $(-B \pm \sqrt{d})/(2A)$, which involve only field operations and the extraction of a single square root. If $\sqrt{d} \in K$, both roots lie in $K$, and the new point contributes no extension: $[K(p_1, p_2) : K] = 1$. If $\sqrt{d} \notin K$, then $p_1 \notin K$ and $K(p_1) = K(\sqrt{d})$. To see why $K(p_1) = K(\sqrt{d})$: since $p_1 = (-B + \sqrt{d})/(2A)$ (taking one choice of sign), we have $\sqrt{d} = 2Ap_1 + B \in K(p_1)$, so $K(\sqrt{d}) \subset K(p_1)$. Conversely, $p_1$ is expressed in terms of $\sqrt{d}$ and elements of $K$, so $p_1 \in K(\sqrt{d})$, giving $K(p_1) \subset K(\sqrt{d})$. Therefore $K(p_1) = K(\sqrt{d})$. The polynomial $t^2 - d \in K[t]$ is irreducible over $K$ precisely because $d$ has no square root in $K$: if $t^2 - d = (t - e)(t + e)$ for some $e \in K$, then $e^2 = d$, contradicting the assumption $\sqrt{d} \notin K$. By the [Structure of Simple Algebraic Extensions](/theorems/1251), the degree of the simple extension $K(\sqrt{d})/K$ equals the degree of the minimal polynomial of $\sqrt{d}$ over $K$, which is 2. Finally, $p_2 = -(\alpha p_1 + \gamma)/\beta$ is a $K$-linear function of $p_1$ (since $\alpha, \gamma, \beta \in K$), so adjoining $p_2$ to $K(p_1)$ adds nothing: $K(p_1, p_2) = K(p_1)$.[/guided]

[step:Case 3: Intersection of two circles reduces to the line-circle case]Suppose $P$ is an intersection of two distinct circles, both determined by points of $S$. The coordinates $(p_1, p_2)$ satisfy \begin{align*} x^2 + y^2 + \delta_1 x + \varepsilon_1 y + \zeta_1 &= 0, \\ x^2 + y^2 + \delta_2 x + \varepsilon_2 y + \zeta_2 &= 0, \end{align*} with $\delta_i, \varepsilon_i, \zeta_i \in K$ for $i \in \{1, 2\}$, and the two circles are distinct. Subtract the second equation from the first: \begin{align*} (\delta_1 - \delta_2) x + (\varepsilon_1 - \varepsilon_2) y + (\zeta_1 - \zeta_2) = 0. \end{align*} This is a linear equation in $x$ and $y$ with coefficients in $K$. Define $\alpha := \delta_1 - \delta_2$, $\beta := \varepsilon_1 - \varepsilon_2$, and $\gamma := \zeta_1 - \zeta_2$, all in $K$. We claim $(\alpha, \beta) \neq (0, 0)$. If $\alpha = 0$ and $\beta = 0$, then $\delta_1 = \delta_2$ and $\varepsilon_1 = \varepsilon_2$, and the subtracted equation gives $\zeta_1 = \zeta_2$ (since $P$ satisfies the equation, so $\gamma = \zeta_1 - \zeta_2 = 0$). But then the two circle equations are identical, contradicting the hypothesis that the circles are distinct. The line $\alpha x + \beta y + \gamma = 0$ is called the **radical axis** of the two circles. Every intersection point of the two circles lies on this line. In particular, $P$ satisfies both the radical axis equation and the first circle equation. This is precisely a line-circle intersection with coefficients in $K$. By Case 2, $[\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] \in \{1, 2\}$.[/step]

[guided]The algebraic trick for the circle-circle case is to subtract one circle equation from the other. Since both equations have the form $x^2 + y^2 + (\text{linear and constant terms})$, the quadratic terms $x^2 + y^2$ cancel, leaving a linear equation — the equation of the radical axis. Why does this reduction work? A point $P = (p_1, p_2)$ lying on both circles satisfies both equations simultaneously. If $P$ satisfies both $x^2 + y^2 + \delta_1 x + \varepsilon_1 y + \zeta_1 = 0$ and $x^2 + y^2 + \delta_2 x + \varepsilon_2 y + \zeta_2 = 0$, then $P$ satisfies their difference, which is the radical axis equation. Therefore the system "two circles" is equivalent to the system "radical axis plus one circle" — a line-circle system. The only subtlety is verifying that the radical axis equation is non-degenerate, i.e., $(\alpha, \beta) \neq (0, 0)$. If $\alpha = \delta_1 - \delta_2 = 0$ and $\beta = \varepsilon_1 - \varepsilon_2 = 0$, then the two circles have the same centre (expanding the general circle equation $x^2 + y^2 + \delta x + \varepsilon y + \zeta = 0$, the centre is $(-\delta/2, -\varepsilon/2)$). For two concentric circles, the subtracted equation becomes $\zeta_1 - \zeta_2 = 0$, which is either always true (if $\zeta_1 = \zeta_2$, meaning the circles are identical, contradicting distinctness) or never true (if $\zeta_1 \neq \zeta_2$, meaning the circles are concentric with different radii and do not intersect, contradicting the existence of $P$). In either case, we reach a contradiction, so $(\alpha, \beta) \neq (0, 0)$. Having verified the non-degeneracy, we apply Case 2 to the line-circle system and conclude $[\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] \in \{1, 2\}$.[/guided]

[step:Conclude that all cases give degree 1 or 2] Every 1-step constructible point $P$ arises from one of the three cases analysed above: line-line, line-circle, or circle-circle intersection. In Case 1, $[\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] = 1$. In Cases 2 and 3, $[\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] \in \{1, 2\}$. Therefore \begin{align*} [\mathbb{Q}(S \cup \{P\}) : \mathbb{Q}(S)] \in \{1, 2\}. \end{align*} [/step]