Prove that $Q$ is rational if and only if $t$ is rational In the accompanying diagram from "The Language of Mathematics: Making the Invisible Visible" by Keith Devlin (Figure 6.21), he states that "It is then an easy exercise in algebra and geometry to verify that the point $Q$ will have rational coordinates if, and only if, the number $t$ is rational."  Note that the circle is the unit circle, and so the coordinates of $P$ are $(-1,0)$.

He gives certain examples, such as if $t = 1/2$, "a little computation shows that the point $Q$ has coordinates $(3/5, 4/5)$. Similarly, $t = 2/3$, leads to the point $(5/13, 12/13)$."
I understand these coordinates yield Pythagorean Triplets, but how are they obtained algebraically (and/or geometrically)? and how is the general case he gives above—regarding $Q$ being rational if and only if $t$ is rational—proven?
 A: The equation of the line is
$$y=t(x+1)$$
and the circle's,
$$x^2+y^2=1$$
The first equation implies that $x,y\in\Bbb Q\implies t\in\Bbb Q$.
Merging both of them we obtain
$$x^2+t^2(x+1)^2=1$$
 But if $t$ is rational, this is a quadratic equation on $x$ with rational coefficients and one rational root (namely, $-1$). So the other solution is rational, too.
 This implies that $t\in\Bbb Q\implies x,y\in\Bbb Q$
A: It follows from your post, but you didn't define it, that that is the unit canonical circle $\;x^2+y^2=1\;$ and thus $\;P(-1,0)\;$. Call $\;(0,t)\;$ the point where that cord intersects the $\;y\,-$ axis, that so by basic trigonometric we have that the slope of $\;PQ\;$ is  just $\;t\;$ , and if we call $\;Q(a,b)\;$ the coordinates of $\;Q\;$ , then
$$t=m_{QP}=\frac b{a+1}\implies b=t(a+1)$$
and since $\;Q\;$ on the unit circle, we get:
$$1=a^2+b^2=a^2+a^2t^2+2at^2+t^2\implies(1+t^2)a^2+2t^2a+t^2-1=0$$
The discriminant of the above quadratic in $\;a\;$ is
$$\Delta=4t^4-4(t^2+1)(t^2-1)=4\implies a_{1,2}=\frac{-2t^2\pm2}{2(t^2+1)}=\begin{cases}-1\\{}\\-\cfrac{t^2-1}{t^2+1}\end{cases}$$
Now just prove that the above means $\;a\;$ (and thus also $\;b\;$) are rational iff $\;t\;$ is...
A: A high-level handwavy explanation of why this must be so:
The $x$-coordinates of the two intersection points between the line and the circle are solutions to a quadratic equation whose coefficients can be computed from the known points $(-1,0)$ and $(0,t)$ by purely rational operations.
Thus, the quadratic equation has rational coefficients. Such a quadratic equation has either both roots rational or both roots irrational. (This is clear from the quadratic formula).
Since $P$ has a rational $x$-coordinate, $Q$ therefore also has.
Similarly for the $y$ coordinates.
A: The line $PQ$ is given by
$$
y=tx+t
$$
Inserting this into the circle equation
$$
x^2+y^2=1
$$
we get
$$
x^2+(tx+t)^2=1\\
(t^2+1)x^2+2t2x+t^2-1=0\\
x^2+\frac{2t^2}{t^2+1}x+\frac{t^2-1}{t^2+1}=0
$$
This is the equation that gives the $x$-coordinates of $P$ and $Q$. However, we already know the $x$-coordinate of $P$. So Vieta's formulas (or polynomial division, or actually solving the quadratic) tells us that the $x$-coordinate of $Q$ is $\frac{1-t^2}{1+t^2}$. Using the equation for the line, that gives a $y$ coordinate of
$$
t\cdot\frac{1-t^2}{1+t^2}+t=\frac{t-t^3+t+t^3}{1+t^2}\\
=\frac{2t}{1+t^2}
$$
So, clearly, if $t$ is rational, the $x$ and $y$ coordinates of $Q$ are both rational.
On the other hand, if the $x$ and $y$ coordinates of $Q$ are both rational, then the slope of the line (using the coordinates of $P$ and $Q$ to calculate it) is rational. And this slope is $t$.
A: The function $f(x)=\dfrac{1-x}{1+x},\; x\neq -1$ is its own inverse.
The slope of the line $PQ$ is $t$ and also $\dfrac y{1+x}=\sqrt{\dfrac{1-x}{1+x}}$. 
Then, it is clear that if $x,y$ are rational, then $t$ either is. 
Also, $t^2 =f(x)$, and hence $f(t^2)=x$. Then, if $t$ is rational, $x$ and $y$ either are.
