Generating Pythagorean triples for $a^2+b^2=5c^2$?

Just trying to figure out a way to generate triples for $a^2+b^2=5c^2$. The wiki article shows how it is done for $a^2+b^2=c^2$ but I am not sure how to extrapolate.

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This is one of those CW answers. Country and Western.

I did Gerry’s recipe and I quite like how it works. Educational, you might say.
I took the slope $t = \frac{q}{r}$ and starting rational point $(2,1).$ The other point works out to be
$$ x = \frac{2 t^2 – 2 t – 2}{t^2 + 1}, \; \; \; y = \frac{- t^2 – 4 t + 1}{t^2 + 1}, $$ so multiply everything by $r^2$ to arrive at

$$ x = \frac{2 q^2 – 2qr – 2r^2}{q^2 + r^2}, \; \; \; y = \frac{- q^2 – 4 qr + r^2}{q^2 + r^2}. $$
So far $x^2 + y^2 = 5.$ Multiply through by $q^2 + r^2$ to get

$$ a = 2 q^2 – 2 q r – 2 r^2 $$

$$ b = -q^2 – 4 q r + r^2 $$

$$ c = q^2 + r^2 $$
$$ a^2 + b^2 = 5 q^4 + 10 r^2 q^2 + 5 r^4 $$
and
$$ c^2 = q^4 + 2 r^2 q^2 + r^4 $$
and
$$ a^2 + b^2 = 5 c^2 $$

Consider the circle $$x^2+y^2=5$$ Find a rational point on it (that shouldn’t be too hard). Then imagine a line with slope $t$ through that point. It hits the circle at another rational point. So you get a family of rational points, parametrized by $t$. Rational points on the circle are integer points on $a^2+b^2=5c^2$.