EDIT1:
For positive power Circles two variable segments $OP,OQ$ at any inclination originating from an external point $O $ shown at left we can draw constant Arithmetic mean $ OA=\frac{OP+OQ}{2}$, Geometric mean $ OG=\sqrt{OP.OQ}$ and Harmonic mean $ OH=\frac{2.OP.OQ}{OP+OQ}$ represented by Circles in magenta, green and blue respectively.
Correspondingly for negative power how can we draw constant AM,GM and HM circles (if at all the locus is a circle) from inside point $O$ at right to find the $A,G,H$ curves?
Thanking you in advance.
Without loss of generality, by translational, rotational and scaling symmetries, we can assume that $O=(0,0)$, the power line makes an angle $\alpha$ with respect to the x-axis and the circle has radius $1$ and it's center is at $C=(c,0),0\leq c\leq1$. Then $P,Q$ are the solutions to $$\begin{cases}x\sin\alpha-y\cos\alpha=0\\(x-a)^2+y^2=1\end{cases}\Rightarrow\begin{cases}y=x\tan\alpha\\x^2-2ax+a^2+x^2\tan^2\alpha=\displaystyle\frac{x^2}{\cos^2\alpha}-2ax+a^2=1\end{cases}$$ $$x=\frac{2a\pm\sqrt{4a^2-4\frac{a^2-1}{\cos^2\alpha}}}{\frac{2}{\cos^2\alpha}}=\cos^2\alpha\left(a\pm\sqrt{\frac1{\cos^2\alpha}-a^2\tan^2\alpha}\right)$$ $$y=x\tan\alpha=\cos\alpha\sin\alpha\left(a\pm\sqrt{\frac1{\cos^2\alpha}-a^2\tan^2\alpha}\right)$$ Then, the distances to O are $$\sqrt{x^2+y^2}=\left\lvert\cos\alpha\left(a\pm\sqrt{\frac1{\cos^2\alpha}-a^2\tan^2\alpha}\right)\right\rvert$$ Let's see what sign we must take $$a^2\geq\frac1{\cos^2\alpha}-a^2\tan^2\alpha\Rightarrow\frac{a^2}{\cos^2\alpha}\geq\frac1{\cos^2\alpha}\Rightarrow a^2\geq1\qquad!!$$ Then $$\sqrt{x^2+y^2}=\lvert\cos\alpha\rvert\left(\sqrt{\frac1{\cos^2\alpha}-a^2\tan^2\alpha}\pm a\right)=\sqrt{1-a^2\sin^2\alpha}\pm a\lvert\cos\alpha\rvert$$ Now we have two options in the signs of $OP,OQ$. Geometrically, they are opposite, but the GM and the HM are only well defined for non-negative numbers. Let's try both options.
If we consider both distances to be possitive, for the AM we have: $$AM=\frac{OP+OQ}2=\sqrt{1-a^2\sin^2\alpha}$$ $$\overrightarrow{AM}=AM\widehat{OP}=\sqrt{1-a^2\sin^2\alpha}\ (\cos\alpha,\sin\alpha)$$ $$y=\sin\alpha\sqrt{1-a^2\sin^2\alpha}\Rightarrow y^2=\sin^2\alpha-a^2\sin^4\alpha$$ $$\sin^2\alpha=\frac{1\pm\sqrt{1-4a^2y^2}}{2a^2}$$ $$x=\cos\alpha\sqrt{1-a^2\sin^2\alpha}\Rightarrow x^2=(1-\sin^2\alpha)(1-a^2\sin^2\alpha)=1-(a^2+1)\sin^2\alpha+a^2\sin^4\alpha=$$ $$=1-y^2-a^2\sin^2\alpha=1-y^2-\frac{1\pm\sqrt{1-4a^2y^2}}2$$ $$2(x^2+y^2)-1=\sqrt{1-4a^2y^2}\Rightarrow 4(x^2+y^2)^2-4(x^2+y^2)+1=1-4a^2y^2$$ $$(x^2+y^2)^2=x^2+(1-a^2)y^2$$ Which is a hippopede.
For the GM we have: $$GM=\sqrt{OP\cdot OQ}=\sqrt{1-a^2\sin^2\alpha-a^2\cos^2\alpha}=\sqrt{1-a^2}$$ And so, it form the circle $x^2+y^2=1-a^2$, which has as diameter the perpendicular to $OC$ at $O$.
For the HM we have: $$HM=\frac{2OP\cdot OQ}{OP+OQ}=\frac{GM^2}{AM}=\frac{1-a^2}{\sqrt{1-a^2\sin^2\alpha}}$$ $$\overrightarrow{HM}=HM\widehat{OP}=\frac{1-a^2}{\sqrt{1-a^2\sin^2\alpha}}(\cos\alpha,\sin\alpha)$$ $$y=\sin\alpha\frac{1-a^2}{\sqrt{1-a^2\sin^2\alpha}}\Rightarrow y^2=\sin^2\alpha\frac{(1-a^2)^2}{1-a^2\sin^2\alpha}\Rightarrow\sin^2\alpha=\frac{y^2}{(1-a^2)^2+a^2y^2}$$ $$x=\cos\alpha\frac{1-a^2}{\sqrt{1-a^2\sin^2\alpha}}\Rightarrow x^2=(1-\sin^2\alpha)\frac{(1-a^2)^2}{1-a^2\sin^2\alpha}=(1-a^2)^2\frac{1-\frac{y^2}{(1-a^2)^2+a^2y^2}}{1-a^2\frac{y^2}{(1-a^2)^2+a^2y^2}}=$$ $$=(1-a^2)^2\frac{(1-a^2)^2+a^2y^2-y^2}{(1-a^2)^2+a^2y^2-a^2y^2}=(1-a^2)^2\left(1-\frac{y^2}{1-a^2}\right)$$ $$\frac{x^2}{(1-a^2)^2}+\frac{y^2}{1-a^2}=1$$ Which is an ellipse.
This leaves the following curves.
If we consider one distance to be negative, then we have for the AM: $$AM=\frac{OP+OQ}2=a\lvert\cos\alpha\rvert$$ $$\overrightarrow{AM}=AM\cdot\widehat{OP}=a\lvert\cos\alpha\rvert(\lvert\cos\alpha\rvert,\sin\alpha)$$ $$x=a\lvert\cos\alpha\rvert^2=a\cos^2\alpha$$ $$y=a\lvert\cos\alpha\rvert\sin\alpha\Rightarrow y^2=a^2\cos^2\alpha(1-\cos^2\alpha)=ax\left(1-\frac xa\right)=ax-x^2$$ $$\left(x-\frac a2\right)^2+y^2=\frac{a^2}4$$ Which is the circle with diameter $OC$.
The GM doesn't exist because is the square root of a negative number. For the HM we have: $$HM=\frac{2OP\cdot OQ}{OP+OQ}=\frac{a^2-1}{a\lvert\cos\alpha\rvert}$$ $$\overrightarrow{HM}=AM\cdot\widehat{OP}=\frac{a^2-1}{a\lvert\cos\alpha\rvert}(\lvert\cos\alpha\rvert,\sin\alpha)$$ $$x=\frac{a^2-1}{a}=a\cos^2\alpha$$ Which is a straight line.