Show that $AT \cdot XY = AX \cdot BY$ In figure below $T,X,Y$ are tangency points and the big circles are tangent to each other. And $TB$ is parallel to that common tangent below.
So, I made more calculations than I should. Let $R$ be the radius of the blue $\Omega$ circle and $r$ be the radius of the green circle.
I managed, through heavy calculation, to come to:
$R = r + \frac{AB^2}{16r}$
I guess this is a good start, there should be an easy way to prove it.
By defining $TA = t, AX = a, XY = 2k, BY = b$
I also came to the following relation:
$(4t^2 + 4t(a+b+2k) + (a+b+2k)^2)\frac1{\sqrt{8t^2 + 8t(a+b+2k) + (a+b+2k)^2}} k = k^2 + (a+k)(b+k)$
which is nice, has the same variables that the question asks, can anyone solve this one?
My calculations:
$M =$ midpoint of $\overline{AB}$
$C = \Omega \cap$ the common tangent, $CA=CB$
$D$ is such that $TMCD$ is a rectangle and $E$ is the tangency point between $\Omega$ and green circle. $TE = x \implies EC = \frac{R}{r} x$ (homothety centered at $E$)
$TC^2 = TD^2 + DC^2 = (2r)^2 + (2\sqrt{Rr})^2 = (x\frac{R+r}r)^2$
so $x = \frac{2r}{R+r} \sqrt{r(r+R)}$ RIGHT?
$x \cdot TC = TA \cdot TB \implies 4r^2 = TA \cdot TB$ RIGHT?
but $TB = 2\sqrt{Rr} + \frac{AB}2$
So now:
$TA = \frac{\sqrt{16^2-AB^2}-AB}2$ oops, I think I see my problem.... I tried to isolate $R$ in $TA \cdot (2\sqrt{Rr} + \frac {AB}2) = 4r^2$ and I changed a signal.... welp if someone wants to continue from here.
 A: The answer is written before the edit to your question...
Let $R$ be the radius of $\Omega$, $p$ be the distance of the center of $\Omega$ to $TB$. The circle tangent at $X$ and $Y$ have radius $r_1$ and $r_2$, and the centers of whom are $U,V$. The center of $\Omega$ is $W$.
So we have $UW=R-r_1$, the vertical distance of $U$ and $W$ is $r_1+p$, so $$AX=\sqrt{R^2-p^2}-\sqrt{(R-r_1)^2-(p+r4_1)^2}=\sqrt{R+p}(\sqrt{R-p}-\sqrt{R-p-2r_1}).$$ Similarly, we have $$YB=\sqrt{R+p}(\sqrt{R-p}-\sqrt{R-p4-2r_2}).$$ Also, $XY$ is the sum of two horizontal distances, that is, $$\sqrt{(R-r_1)^2-(p+r_1)^2}+\sqrt{(R-r_2)^2-(p+r_2)^2}=\sqrt{R+p}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2}).$$ Also we know that this $XY$ also equals to $\sqrt{4r_1r_2}$. So we have the relation
$$\sqrt{R+p}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=\sqrt{4r_1r_2}$$
Also, the radius of the green circle is $\frac{R+p}{2}$. So the common tangent of the green circle and $\Omega$ is $\sqrt{2R(R+p)}$, and thus $AT=\sqrt{2R(R+p)}-\sqrt{R^2-p^2}=\sqrt{R+p}(\sqrt{2R}-\sqrt{R-p})$. So we have
$$AT\times XY=AX\times BY$$
is equivalent to
$$(\sqrt{2R}-\sqrt{R-p})(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=(\sqrt{R-p}-\sqrt{R-p-2r_1})(\sqrt{R-p}-\sqrt{R-p-2r_2})$$
That is, we have
$$\sqrt{2R}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=R-p+\sqrt{R-p-2r_1}\sqrt{R-p-2r_2}$$
Let $x=\sqrt{R-p-2r_1}$, $y=\sqrt{R-p-2r_2}$, we want to prove that
$$\sqrt{2R}(x+y)=R-p+xy$$
Since we have the relation
$$\sqrt{R+p}(\sqrt{R-p-2r_1}+\sqrt{R-p-2r_2})=\sqrt{4r_1r_2}$$
That is,
$$\sqrt{R+p}(x+y)=\sqrt{(R-p-x^2)(R-p-y^2)}$$
Square it, we have
$${(R+p)}(x+y)^2=(R-p-x^2)(R-p-y^2)$$
Or,
$${(R+p)}(x+y)^2=(R-p)^2-(R-p)(x+y)^2+2(R-p)(xy)+(xy)^2$$
Move the term, we have
$${2R}(x+y)^2=(R-p)^2+2(R-p)(xy)+(xy)^2$$
Or,
$${2R}(x+y)^2=(R-p+xy)^2$$
take the square root, we have
$$\sqrt{2R}(x+y)=R-p+xy$$
Which yields the desired result.
A: Besides $R$ and $r$, let $h$ be the distance from the center of the large circle to $AB$ and let $AB = 2l,$ for simplicity. Since the two lines are parallel $R+h = 2r$ and you can find $AT$ by Pythagora:
$$(AT+l)^2 + (R-r)^2 = (R+r)^2\implies AT = 2\sqrt{Rr} - l.$$
If you let $AX = a$ and $BY = b$, then it suffices to show that:
$$\dfrac{AX\cdot BY}{XY} = \dfrac{ab}{2l-a-b} = AT = 2\sqrt{Rr} - l.$$
First step is to find an equation relating $a$ and $b.$ Let $x$ be the radius of the small circle corresponding to $a$, then another Pythagora's theorem says:
$$(l-a)^2 + (h+x)^2 = (R-x)^2\implies x  = \dfrac{R^2-h^2 - (l-a)^2}{2(R+h)} = \dfrac{a(2l-a)}{4r}.$$
This is because:
$$R+h = 2r\quad \text{ and }\quad R^2- h ^2 = l^2.$$
Similarly,  if $y$ is the radius of the small circle corresponding to $b$, then $y = \dfrac{b(2l-b)}{4r}.$ Finally, if you do another Pythagora's theorem on the two small circles being tangent to each other:
$$XY^2 = (x+y)^2 - (x-y)^2 = 4xy\implies (2l-a-b)^2 = \dfrac{ab(2l-a)(2l-b)}{4r^2}.$$
You can manipulate this easily into an equivalent form:
$$\dfrac{ab}{2l-a-b} = \dfrac{4r^2}{2l + \dfrac{ab}{2l-a-b}}\implies \dfrac{ab}{2l-a-b} = -l + \sqrt{l^2+4r^2} = -l + \sqrt{R^2 - h^2 + 4r^2} = $$
$$ = -l + \sqrt{R^2-(R-2r)^2+4r^2} = -l+2\sqrt{Rr},$$
as desired.
