Why is the ratio $\overline{HO}/\overline{HM}$ constant for all $\angle IHG$? In the first attached diagram, the smaller circle is centered on point $H$, and the larger circle is centered on point $G$. The long line connecting points $L$ and $K$ is the perpendicular bisector of segment $\overline{IH}$. Since $H$ is the center of the smaller circle, $M$ is the midpoint between $L$ and $K$. Finally, the point $O$ is the circumcircle center of triangle $LKG$.
I notice in Geogebra that the ratio $\overline{HO}/\overline{HM}$ is constant for any angle $\angle IHG$. Why? I cannot come up with a proof. It's starting to drive me crazy.
To be clear, the focal distance $\overline{HG}$ and circle radii are held constant as angle $\angle IHG$ varies.
It might also be useful to consider (second attached diagram) that the intersection of the radius $\overline{IG}$ with segment $\overline{LK}$ is a (pink) point $P$ on the blue ellipse with focii $H$ and $G$.


 A: (It's ever-so-slightly cleaner to consider $|HO|/|HI|$, so I'll do that.)

Let $g$, $h$, and $r$ be the radii of $\bigcirc G$, $\bigcirc H$, and $\bigcirc O$; and define
$$p := |HI| \qquad q := |HO| \qquad d:=|HG|$$
Note that $\square HKIL$ is a rhombus, so that $|LI|=h$.

We see that $\triangle HGI$ and $\triangle HLI$ have congruent cevians to $O$ on their common base $\overline{HI}$. By Stewart's Theorem, we have
$$d^2(p-q)+g^2q\;\underbrace{=}_{\triangle HGI}\;\;\underbrace{p\,(r^2+q(p-q))}_{\text{(doesn't matter)}}\;\;\underbrace{=}_{\triangle HLI}\;h^2(p-q)+h^2q \tag1$$
Re-writing gives
$$\frac{q}{p}=\frac{d^2-h^2}{d^2-g^2} \tag{$\star$}$$
which is a constant for the configuration. $\square$
A: 
In left figure the radius of little circle s half of that of big one. in this case $\frac{HO}{HM}=1$.
In right figure we compare position of HI in vertical position with that of in horizontal position. As can be seen the locus of points M and O are ellipses with common center H when angle between HI and HG varies , that is why the ratio $\frac {HO}{HM}$ is constant.
