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Consider the unit circle

$$ x^2+y^2=1. $$

How can I parametrize it using the line $y=m(x+1)$, where $m$ is its slope?

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up vote 5 down vote accepted

For the standard parametrization, calculate the second point where the line meets the circle. This is the solution of the system $x^2+y^2=1$, $y=m(x+1)$. Substitute for $y$ in the equation of the circle. We get $x^2+(m(x+1))^2=1$, which expands to $$(1+m^2)x^2 +2mx +m^2-1=0.$$ The equation has the obvious root $x=-1$, since $(-1,0)$ lies on both the line and the circle. The product of the roots is $\dfrac{m^2-1}{m^2+1}$. Thus the other root of the quadratic is $\dfrac{1-m^2}{1+m^2}$.

That tells us that $$x=\frac{1-m^2}{1+m^2}.$$ We also need $y$ in terms of the parameter $m$. This is not difficult now that we know $x$.

Remark: We don't quite cover the full circle, for $x=-1$ does not correspond to any real value of the parameter $m$. But here is a use for "$\infty$" The point $(-1,0)$ is taken care of by letting $m=\infty$.

A nice feature of the above parametrization is that our point $(x,y)$ has rational coordinates iff $m$ is rational. That gives us a nice way to describe all the rational solutions of the equation $x^2+y^2=1$.

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