Every Pythagorean triple

Call positive integers $(a,b,c)$ a primitive Pythagorean triple if $a^2+b^2=c^2$ and $\gcd (a,b,c)=1$. Such a triple has exactly one even leg; call it $b$.

Show that every primitive triple has the form
$$a=m^2-n^2,b=2mn,c=m^2+n^2$$
for coprime integers $m>n>0$ of opposite parity, and that every such pair $(m,n)$ gives one.

There is an algebraic proof by factoring, and a geometric one that draws lines through a point on a circle.