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I am doing something with the curve given parametrically by

$y = (-ar+b) r$, $x = \sqrt{r^2-y^2}$

for $r\in \lbrack (b-1)/a,b/a\rbrack$. It is nice enough (and of low enough degree) that I suspect it has been studied before - and, in particular, that it has a name. Does it?

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Eliminating $r$ gives the equation $-b^2 (x^2 + y^2) + (y + a (x^2 + y^2))^2=0$. – Mariano Suárez-Alvarez Jan 3 '13 at 17:49
up vote 11 down vote accepted

I'm assuming that $a$ is nonzero, in which case, this curve of degree $4$ is a rational curve: Just set $x=r\cos\theta$ and $y=r\sin\theta$, then you get $\sin\theta = b-ar$, so setting $$ \cos\theta = \frac{1-t^2}{1+t^2}\qquad\qquad{\text{and}}\qquad\qquad \sin\theta = \frac{2t}{1+t^2} = b - a r, $$ you can now solve for r as a function of $t$ and thence get $x$ and $y$ as rational expressions in $t$. The $t$-range corresponding to your given $r$-range is $0\le t\le 1$.

It turns out that this is a limaçon, as the equation $r = (b-\sin\theta)/a$ is the classical equation for a limaçon in polar coordinates.

When $a=0$, the curve is a line (segment).

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Ah! Good catch :-) The cartesian equation given at is the one I got by elimination in the comment above. – Mariano Suárez-Alvarez Jan 3 '13 at 20:28
Nice! This is the reference I needed. – H A Helfgott Jan 5 '13 at 22:05

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