It is known that the if the product of two line segments OA,OB drawn from any point O to a curve is a constant, then the curve is a circle (black). The product is the square of the geometric mean.
However what is another curve (green) if harmonic mean of OA,OB is a constant?
Thanks for all suggested insights or the curve itself if already known.
On
A solution obtained when O,A,B are not always collinear, even if derived on that basis. Could be due to some inherent precession? in the spiral.
A circle radius $a$ is drawn through A,B, centred on their perpendicular bisector. Let
$$OB=r, OA= r- 2 a \sin \psi, \psi= \psi(s), r'(s)= \cos \psi $$
Constant Harmonic mean
$$H=\frac{r (r- 2 a \sin \psi)}{(r- a \sin \psi)} \tag1$$
Differentiate wrt arc $s$ using Chain Rule, cancelling $\cos \psi$ at RHS
$$ H= \frac{2 r ~- 2 a \sin \psi - 2 a r \sin \psi }{1- a \psi'(s)}$$
Cross multiply, simplify to get ode of desired curve
$$ a \psi ' =1 +\frac{2 a \sin \psi}{r} \left(\frac{ a \sin \psi}{r} -1\right) \tag 2$$
From (1)
At tangent point $$ \psi=0, ~ H=r= r_{tangent}$$
At $$~ \psi = \pi/2, ~r_{max}= a+ H/2 + \sqrt{a^{2}+H^{2}} \tag 3$$
At $ \psi =3 \pi/2$
$$\psi =3 \pi/2,~~r_{min}=- a+ H/2 - \sqrt{a^{2}+H^{2}},~r_{max}- r_{min}= 2a\tag 4$$
Also from (1) $$ H= \frac{ r_{max}.(r_{max}-2 a)}{r_{max}-a} = \frac{ r_{min}.(r_{min}+2 a)}{r_{min}+a} ; \tag5 $$
The spiral drawn to input $ a=1, H=1$ results in extreme radii (1,3) calculated by above relations in the advancing spiral:
I have understood that you would like to find a type of curve such that, whatever the pole $O$ we have, for any secant line issued from $O$ :
$$\text{HarmMean(OA,OB)=constant}$$
What I am giving below is a solution restricted to the case of a fixed pole (taken as the origin) ; nevertheless, it is a "piece of the puzzle" helping in the direction of finding either a general solution... or a proof that such a general type of curve doesn't exist.
The particular solution I propose (a self-invariant curve with respect to its center)is in fact the union of two curves with resp. polar equations :
$$\begin{cases}r_1(t)&=&\frac{\tan(t)}{1+\tan(t)}&&& (0 \le t \le 3 \pi/4 \ \ \text{blue curve})\\ r_2(t)&=& \frac{\tan(t)}{\tan(t)-1}&=&-\frac{r_1(t)}{2r_1(t)-1} &(\pi/4 \le t \le \pi\ \ \text{red curve})\end{cases}$$
(the two line asymptotes are not part of the curve).