Googly Eyed Triangle

$\text{P and R are the centers, }~~~~ RS\perp PX,~~ YQ \perp XZ . \\SR=\sqrt{(2+1)^2-1^2}=\sqrt 8,~~\therefore ~XQ=\dfrac 2 3 \sqrt 8 , ~~ZQ=\dfrac 1 3 \sqrt 8.\\YQ=\dfrac 1 3 PS +SX=\dfrac 4 3 .\\XY^2+YZ^2=(XQ^2+YQ^2) + (ZQ^2+YQ^2) =~~~~~\large \color{#D61F06}{8}$

Let $E$ be the center of $\omega _1$ and $F$ be the center of $\omega _2$ . Let $G$ be the foot of the perpendicular dropped from $F$ onto $EX.$ We have $FG = XZ$ because $FGXZ$ is a rectangle. Also, $EG = 1$ and $FE = 3.$ Solving for $FG$ using right triangle $EFG$ yields $FG = XZ = \sqrt{8}.$

Now, from Power of a Point, we deduce that $XD = DY = DZ.$ This shows that $XYZ$ is a right triangle, which can be seen by drawing a circle with center $D$ passing through $X.$ Thus, $XY^2 + YZ^2 = \sqrt{8}^2 = \boxed{8}.$

By LoC, note that we have $XY^2 = 2r^2 + 2r^2 \cos \theta$ and $YZ^2 = 2r^2-2r^2 \cos \theta$ , giving their sum to be $4r^2$ where r is the tangent length. By Equal Tangents, we note that $XD = ZD = r$ . By Pythagoras, we have that $XZ = 2\sqrt{2}$ , giving the answer as $8$ .