Planar curves identical to their inverses

Question

Is the right strophoid the only planar curve $C$ whose inverse curve w.r.t. some circle (in this case: centered on the origin) is identical to $C$?

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          ^{(Image link.)}

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Answer 1

Why should it be unique? Let us use complex numbers. Let the equation of the curve be $$F(z,\overline{z})=\sum_{i,j}a_{i,j}z^i\overline{z}^j=0.$$ Inversion (with respect to the unit circle) as defined in the reference you give is $z\mapsto 1/\overline{z}$. We obtain the image curve $$\sum_{i,j}a_{i,j}\overline{z}^{a-i}z^{b-j}=0.$$ where $a$ and $b$ are degrees of $F$ with respect to the first and second variable, So the condition is $a_{i,j}=\overline{a_{a-i,b-j}}.$

Remark. According to the definition in your reference, inversion is defined with respect to a circle, not with respect to a point.

Answer 2

This question has been answered but the following (really a comment but not entitled) might add enlightenment. It is easy to construct such curves geometrically by taking one inside the unit circle with both endpoints on the circle and amalgimating it with its inverse. Additional conditions can be imposed to ensure smoothness of the resulting curve. An analytic version is as follows: let $I$ and $J$ be disjoint intervals, $f$ and $g$ suitable functions thereon. Assume that for each $t$ in $I$, there is an $s$ in $J$ with $f(t)= \dfrac 1 {g(s)}$ and vice versa. Then the curve $r=f(\theta)$ for $\theta \in I$ resp. $r=g(\theta)$ for $\theta \in J$ works. Again, any desired smoothness can be obtained by imposing the obvious conditions on the functions.