The answer is $\boxed{Rr}$. We employ coordinates. Set the origin at $O$ and the $x$-axis parallel to $BC$. Clearly it's also parallel to $XY$. Set $I=(u,v)$, $X=(-a,h)$, and $Y=(a,h)$. Note that $a^2+h^2=R^2$ and $OI^2 = u^2+v^2$. [asy][asy] size(6cm); pair A = dir(120), B = dir(210), C = dir(-30), I = incenter(A, B, C), O = origin, Y = dir(30), X = dir(150), O1 = circumcenter(X, I, Y), N = foot(I, X, Y); draw(incircle(A, B, C)^^unitcircle, red); draw(arc(O1, abs(O1-I), 170, 10, CCW), orange); draw(X--Y, fuchsia); draw(N--I^^O1--O, heavymagenta); string[] names = {"$I$", "$O$", "$X$", "$Y$", "$O_1$", "$N$"}; pair[] points = {I, O, X, Y, O1, N}; pair[] ll = {dir(260), dir(40), dir(180), dir(0), O1, N}; for (int i=0; i<names.length; ++i) dot(names[i], points[i], dir(ll[i])); [/asy][/asy] The perpendicular bisector of line $XI$ has equation \[ y - \frac{v+h}{2} = \frac{a+u}{h-v}\left(x-\frac{u-a}{2}\right). \]The perpendicular bisector of line $YI$ has equation \[ y - \frac{v+h}{2} = \frac{-a+u}{h-v}\left(x-\frac{u+a}{2}\right). \]Intersecting these two lines yields that $O_1$ has coordinates \[ O_1 = \left(0, \frac{v+h}{2}+\frac{a+u}{h-v}\cdot\frac{a-u}{2}\right). \]The distance $OO_1$ is simply the $y$-coordinate of $O_1$: \begin{align*} OO_1 &= \frac{(v+h)(h-v)}{2(h-v)} + \frac{(a+u)(a-u)}{2(h-v)} \\ &= \frac{(h^2-v^2)+(a^2-u^2)}{2(h-v)} \\ &= \frac{R^2-OI^2}{2(h-v)} \\ &= \frac{Rr}{h-v}, \end{align*}where we have used Euler's theorem $OI^2=R^2-2Rr$. The distance $NI$ is the difference in the $y$-coordinates of $N$ and $I$: \[ NI = h-v. \]Thus, $OO_1 \cdot NI = Rr$, as needed.

Following lemma kills: Let centers of three non-concentric circles $C_1,C_2,C_3$ lie on $\ell$. Let $O_i$ be the center of $C_i$. Let the radical axis of $C_2,C_3$ cut $\ell$ at $A_1$. Define $A_2,A_3$, similarly. Then, $A_1A_2:A_2A_3=O_1O_2:O_2O_3$. I have a proof which uses linearity of pop (thus quite bashy). Maybe someone has synthetic solution.

Let ray $XI$ meet $\odot (ABC)$ again at $X'.$ At first observe that $$|IN|\cdot |O_1I|=\frac{1}{2} |XI|\cdot |YI|.$$Since $OO_1$ is perpendicular bisector of $XY$ $$\angle YO_1O=\frac{1}{2} YO_1X=\angle YIX',\text{ }\angle O_1OY=\frac{1}{2}\angle XOY=\angle XX'Y=\angle IX'Y.$$Hence $OYO_1\sim X'YI$ and moreover $$\frac{|OO_1|}{|O_1I|}=\frac{|OO_1|}{|O_1Y|}=\frac{|X'I|}{|YI|},$$$$|OO_1|\cdot |IN|=\frac{1}{2}|XI|\cdot |X'I|=\frac{1}{2}(R^2-|OI|^2)=Rr.$$