# y^2 + x^2 = 1

This entry is based on the prologue for the book Elliptic Tales: Curves, Counting, and Number Theory (challenging but quite fun reading on the morning Tube commute!):
$y^2 + x^2 = 1$ is the familiar equation for the unit circle and in the prologue the authors show how a straight line with a rational slope intersects a circle at two point which is rational i.e, of the form $(x, y) = (\frac{a}{b},\frac{c}{d})$ then the second point is also rational and that all such lines trace out the full set of rational points on the circle.

But then the book goes further –

We say that circle $C$ has a “rational parametrization”. This means that the coordinates of every point on the circle are given by a couple of functions that are themselves quotients of polynomials in one variable. The rational parametrization for $C$ we have found is the following:

$(x,y) = (\frac{1 - m^2}{1+m^2},\frac{2m}{1+m^2})$

So this is what I wanted to check… it surely isn’t claiming that all the points on the circle are rational, is it? Merely that the above – if $m$ (which corresponds to the slope of our line through the two points) is rational, generates the full set of rational points on the circle. Because if $m$ is not rational then the second point will not be either? Is that right?

## 2 thoughts on “y^2 + x^2 = 1”

1. Not having read the book, I can only speculate on the intent of the quote, but I think they mean that the rational functions given generate the entire circle, not just the rational points on it. The “rational” in “rational parametrization” may refer to a rational function (quotient of polynomials) rather than a function of a rational argument. The specified function maps all rational values of m to all rational points on the circle and all irrational values of m to all irrational points on the circle.

1. You never let me down! Your explanation makes complete sense, and answers what was confusing me.