Using the radical axis theorem, the lines $\overline{AD}, \overline{BC}, \overline{XY}$ are all concurrent at one point, call it $F$. Now recall by Miquel's theorem in $\triangle FDC$ the fact that quadrilaterals $DAXY$ and $CBXY$ are cyclic implies $FAXB$ is cyclic as well. Denote $\omega_{3}\equiv(FAXB)$ and $Z\equiv\ell\cap\overline{FXY}$.
Since point $Z$ lies on the radical axis of $\omega_{1},\omega_{2}$, it has equal power with respect to both circles, thus \[AZ^{2}=\text{Pow}_{\omega_{1}}(Z)=ZX\cdot ZY=\text{Pow}_{\omega_{2}}(Z)=ZB^{2}\implies AZ=ZB.\] Also, notice that \[AZ\cdot ZB=\text{Pow}_{\omega_{3}}(Z)=ZX\cdot ZF\implies ZY=ZF.\] The diagonals of quadrilateral $FAYB$ bisect each other at $Z$, so we conclude that $FAYB$ is a parallelogram. Let $u:=ZX$, so that $ZY=ZF=u+47$.
Because $FAYB$ is a parallelogram and quadrilaterals $DAXY, CBXY$ are cyclic, \[\angle DFX=\angle AFX=\angle BYX=\angle BCX=\angle FCX~~\text{and}~~\angle XDF=\angle XDA=\angle XYA=\angle XFB=\angle XFC\] so we have the pair of similar triangles $\triangle DFX~\sim~\triangle FCX$. Thus \[\dfrac{37}{2u+47}=\dfrac{2u+47}{67}\implies 2u+47=\sqrt{37\cdot 67}\implies u=\dfrac{1}{2}\left(\sqrt{37\cdot 67}-47\right).\] Now compute \[AB^{2}=4AZ^{2}=4\cdot ZX\cdot ZY=4u(u+47)=37\cdot 67-47^{2}=\textbf{270}.\]
[AIME 2016-I15 Geogebra Diagram.png](https://artofproblemsolving.com/wiki/index.php/File:AIME_2016-I15_Geogebra_Diagram.png)
Circles $\omega_1$ and $\omega_2$ intersect at points $X$ and $Y$. Line $\ell$ is tangent to $\omega_1$ and $\omega_2$ at $A$ and $B$, respectively, with line $AB$ closer to point $X$ than to $Y$. Circle $\omega$ passes through $A$ and $B$ intersecting $\omega_1$ again at $D \neq A$ and intersecting $\omega_2$ again at $C \neq B$. The three points $C$, $Y$, $D$ are collinear, $XC = 67$, $XY = 47$, and $XD = 37$. Find $AB^2$.