Geometric proof about rational numbers

Question

Given,

$$b\in \Bbb Q\implies \exists \;p,q\in \Bbb Z:\:b=\frac pq \wedge q \ne 0 $$

Any line passing through $(0,b)$ and $(1,0)$ such that $\forall \;b \in \Bbb Q$, intersect the unit circle at a rational point $(x,y)$ such that the slope of the line through $(0,b)$ and $(1,0)$ is $m=-\frac pq$. Since we want points on the unit circle we know $x^2+y^2=1$. So we want the rational points $(x,y)$ where boths graphs intersect, hence I tried to set both equations equal to each other $$y-mx-b=x^2+y^2-1$$ $$y+\frac pq \left(x-1\right)=x^2+y^2-1$$ I always run into counter example of irrational $x$ and $y$ having rational squares. Therefore, I wonder if my intuition is wrong.

So instead of trying to find the intersecting rational points on the unit circle and the line when the $y$-intercept is $b \in \Bbb Q$, I tried another approach. Basically, what I'm seeing is that since $x^2+y^2=1$ the hypotenuse $h$ of any triangle of sides $x$ and $y$ is equal to 1. The angle $\phi$ between the $x$-axis and the hypotenuse is $\tan^{-1}\left(\frac b1\right)=\tan^{-1}\left( \frac yx\right)$ from similar triangles. Does it follow that $$\tan\phi = b = \frac yx \in \Bbb Q \implies \exists\;r,s,t,u\in\Bbb Z : x=\frac rs \wedge \frac tu \wedge s,u \ne 0$$ is true? This would mean that $\cos \phi = x \in \Bbb Q:x = \frac rs$ and $\sin \phi = y \in \Bbb Q : y = \frac tu$ and I would claim that $$ (i) \quad \forall \; b \in\Bbb Q \implies \exists \; p,q \in \Bbb Z : q \ne 0 \wedge b = \frac pq = \tan \phi = \left( \frac yx \right)$$ $$(ii) \qquad \exists \; r,s,t,u, \in \Bbb Z : x = \frac rs \wedge y = \frac tu \wedge s,u \ne 0$$ $$\therefore \quad (i) \implies (ii)$$

Is this correct? Does this show that for the infinitely many points on the unit circle, those points $(x, y)$ intersected by a line passing through $(0, b\in\Bbb Q)$ and $(1, 0)$ are rational such that $\forall \;b \in \Bbb Q \implies \left(x = \frac rs, y = \frac tu \right)$? If not, am I on the wrong track?

Answer 1

The easiest proof I can think of follows your intuition, via trigonometry: Let the line make an angle of $\theta$ with the $x$-axis. Then it intersects the circle in two points: $(1, 0)$ itself, and $(-\cos 2\theta, \sin 2\theta)$. We will show that both coordinates are rational. (Don't read any further if you don't want more than that hint—I couldn't tell if you don't want to be spoiled.)

First, we see that $\tan \theta = b$. Then

$$ \cos \theta = \frac{1}{\sqrt{1+b^2}} $$

and

$$ \sin \theta = \frac{b}{\sqrt{1+b^2}} $$

Since $\cos 2\theta = 2\cos^2 \theta-1$ and $\sin 2\theta = 2 \sin \theta \cos \theta$, we have

$$ x = -\cos 2\theta = -\frac{1-b^2}{1+b^2} $$

and

$$ y = \sin 2\theta = \frac{2b}{1+b^2} $$

The above image depicts $0 < b < 1$, though that needn't be the case.

Answer 2

For $b\ne 0:$ A point $(x,y)$ on the line satisfies $x+y/b=1,$ that is $$x=1-y/b.$$ A point $(x,y)$ on the circle satisfies $$|x|=\sqrt {1-y^2}.$$ So if both of these hold then $$(1-y/b)^2=x^2=1-y^2$$ which simplifies to $$-2y/b+y^2/b^2=-y^2.$$ Now this holds when $y=0$ iff $x=1-y/b=1-0/b=1$. For the other point, if $y\ne 0$ and $-2y/b+y^2/b^2=-y^2$ then $$-2/b+y/b^2=-y,$$ which is equivalent to $$y=2b/(b^2+1),$$ implying $$x=1-y/b=(b^2-1)/(b^2+1).$$ If $x,y$ are both rational it is necessary that $b$ is rational because $b\ne 0\implies y=2b/(b^2+1)\ne 0$ which implies (from $x+y/b=1$) that $1/b=(1-x)/y.$