Rational Parametrization of a Circle & Pythagorean Triples

Today, we discuss an interesting topic rarely taught in school: The rational parametrization of a unit circle. That is, how to find the x coordinates and y coordinates of a circle expressed as a rational function? The usual parametrization of a circle is (cos t, sin t).

unit circle

We consider the straight line (l), passing through the point A(1,0) and the point (0,h).

The gradient of this line is \displaystyle m=\frac{h-0}{0-1}=-h.

The y-intercept of this line is h.

Hence the equation of line l is \boxed{y=-hx+h}.

We know the equation of the unit circle is \boxed{x^2+y^2=1}.

By solving the two simultaneous equations (boxed), we get a quadratic formula:

x^2(1+h^2)-2h^2x+h^2-1=0

Solving the above using the quadratic formula gives us \displaystyle x=\frac{h^2-1}{1+h^2}.

Using \boxed{y=-hx+h}, we get \displaystyle y=\frac{2h}{1+h^2}.

Hence, \displaystyle \boxed{ (\frac{h^2-1}{1+h^2},\frac{2h}{1+h^2})} is a parametrization of the unit circle.

We can use this to generate Pythagorean triples! Simply choose a value of h, say, h=8.

Then \displaystyle x=\frac{h^2-1}{1+h^2}=\frac{63}{65}.

\displaystyle y=\frac{2h}{1+h^2}=\frac{16}{65}.

Substituting into \boxed{x^2+y^2=1}, and multiplying by the denominator, we get the Pythagorean Triple 16^2+63^2=65^2 . Interesting?


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One Response to Rational Parametrization of a Circle & Pythagorean Triples

  1. Pingback: Video on Simplices and Simplicial Complexes | Singapore Maths Tuition

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