Is there any circle in the Euclidean plane such that the coordinates of each its points are rational numbers? Is there any circle in the $\Bbb{R} \times \Bbb{R}$ such that coordinates of each its point are rational numbers?
Thanks
I know I have asked an elementary question, but because of I had left Mathematics for $14$ years and I have forgotten everything, but I would ask another related question, how can it be proved by concept of measure of a set in the Euclidean plane of course not by sets in the x-axis and y-axis?
 A: Consider the unit circle centered at the origin. it contains the points $(1,0), (\frac1{\sqrt{2}},\frac1{\sqrt{2}})$. The first of these points is rational, the second irrational.
Scaling the radius by either a rational or irrational takes the two points to $\mathbb{Q}, \mathbb{R\setminus Q}$ or $\mathbb{R\setminus Q}, \mathbb{Q}$ respectively.
Similarly for translating the centre.
A: Consider 1) the intermediate value theorem and 2) the density of the irrationals in the reals.
Let $C = $ the circle centered at $(a,b)$ with radius $r$ then $C = \{(x,y)| (x-a)^2 + (x -b)^2 = r^2\}$.
Now let $f:[0,2\pi) \rightarrow C; f(\theta) = (a + r*\cos \theta, b+r*\sin \theta)$.  That is $f$ maps the incidental angle of a point to the cooresponding point of the circle.
Let $f_x(\theta) = a+r\cos \theta$ is a function mapping the angle to the $x$ coordinate and $f_y(\theta = a + r\sin \theta$ is a function mapping the angle to the $y$ coordinate of the circle.  These are both continuous functions.
$f_x(0) = a + r$ and $f_x(\pi/2) = a$.  So by the intermediate value theorem for any $c \in (a, a+ r)$ there is $\theta \in (0, \pi/2)$ so that $f_x(\theta) = c$.
And as the irrationals are dense in $\mathbb R$ we may choose $c$ so that $c$ is irrational. 
So that means $(f_x(\theta), f_y(\theta)) = (c, b+ r\sin (\theta)) \in C$ and $c $ is not rational.
So no circle exists with only rational coordinates.
Basically, you should develop an intuitive sense that no continuous curve through $\mathbb R^n$ can have only rational or irrational terms.
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In general, to be a "shape" there is some continuous $f:[0,1]\rightarrow shape$ so that $f(t) = (x,y) \in shape$.  For any $t_0 < t_1$ with $f(t_0) = (x_0, y_0)$ and $f(t_1) = (x_1, y_1)$ then as $t$ "passes" from $t_0$ to $t_1$, $x$ "passes" from $x_0$ to $x_1$ and $y$ "passes" from $y_0$ to $y_1$... and they pass through irrational points.
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But is possible to have a shape so that for all $(x,y)$ in the shape either $x$ is rational of $y$ is rational and they are never both irrational.  Example a square with rational corners.
A: Another way to think about it: $\Bbb Q^2$ (the set of all ordered pairs of rational numbers) is countable, as will any subset of this set. However, a circle with center $\langle x_0,y_0\rangle$ and radius $r>0$ can be obtained as the image of the injective function $[0,2\pi)\to\Bbb R^2$ given by $$t\mapsto\bigl\langle x_0+r\cos t,y_0+r\sin t\bigr\rangle.$$ Since $[0,2\pi)$ is uncountable, so is its image, and so its image is not a subset of $\Bbb Q^2.$
A: No. Suppose there is a circle whose points only lie at rational coordinates, with center $(\frac{p}{q}, \frac{r}{s})$ and radius $R$ (they must be rational coordinates- why?), then there is obviously also a point on the circle at $(\frac{p}{q}+\frac{R}{\sqrt{2}}, \frac{r}{s}+\frac{R}{\sqrt{2}})$ by trigonometry. If $R$ is rational*, then obviously the coordinates of that point are irrational.
In addition, we can't make a square whose points are only rational, by the same reasoning. If two points have rational coordinates, the differences in coordinates will be rational also, but the distance between the points can be irrational. Can you take it from there?
Edit In the case of the circle, if $R$ is not a rational multiple of $\sqrt{2}$, then the method holds. If $R$ is a rational multiple of $\sqrt{2}$, choose the point on the circle $(\frac{p}{q}+\frac{R \cdot \sqrt{3}}{2}, \frac{r}{s}+\frac{R}{2})$
