Covering the Euclidean plane with constructible lines and circles

It's the countable union of measure 0 sets, so its measure is 0.

The equation for any such constructible line or circle will have algebraic coefficients. So if $x$ and $y$ are algebraically independent (say, $x=\pi$, $y=e^{\pi}$), then $(x,y)$ cannot be drawn in this way.

Here's a geometric (as opposed to algebraic) argument.

Let $C$ be the set of points of $\mathbb{R}^2$ which lie either on a line through two constructible points or on a circle with a radius whose endpoints are constructible, where by 'constructible' I mean constructible from a fixed finite set of points by ruler-and-compass constructions in a finite number of steps.

We will prove that $C \ne \mathbb{R}^2$.

To see this, we'll define a family of sets $P_n$, for $n \in \mathbb{N}$, inductively as follows:

• Let $P_0$ be the (finite) set of points you start with;
• With $P_n$ defined, let $P_{n+1}$ be the set of points lying at an intersection of two (non-coinciding) lines or circle defined from $P_n$, where a 'line defined from $P_n$' means one passing through two distinct elements of $P_n$, and a 'circle defined from $P_n$' means one with a radius whose endpoints are distinct elements of $P_n$.

Next, for each $n \in \mathbb{N}$, let $C_n$ be the set of points in $\mathbb{R}^2$ which lie on a line or circle defined from $P_n$.

We obtain the set $C$ as the union of all of the sets $C_n$: $$C = \bigcup\limits_{n=0}^{\infty} C_n$$

It now suffices to prove that $C_n$ is nowhere-dense for each $n \in \mathbb{N}$, so that $C \ne \mathbb{R}^2$ by the Baire category theorem (see BCT3 here).

To prove this, it suffices to prove that $P_n$ is finite for each $n \in \mathbb{N}$. Indeed, each line and circle defined from $P_n$ is determined by two elements of $P_n$, so that if $P_n$ is finite, then $C_n$ is a finite union of lines and circles. Since lines and circles are nowhere-dense, it follows that $C_n$ is a finite union of nowhere-dense sets, so is itself nowhere-dense.

Finally, proving $P_n$ is finite for each $n \in \mathbb{N}$ is an easy induction on $n$. Indeed, $P_0$ is finite by assumption, and if $P_n$ is finite, then $P_{n+1}$ is finite since two non-coinciding lines or circles may only intersect at $0$, $1$ or $2$ points, and there are only finitely many lines and circles definable from $P_n$.