Intersection of circles

As points A,P lie on circle Y, $YP=YA, \angle APY=\angle YAP$

Similarly for circle X, $AX=QX, \angle XAQ= \angle XQA$

Since $\angle YAP=\angle XAQ$

$\angle YPA= \angle XQA$

Hence quadrilateral $YXQP$ is cyclic.

$\angle XPG=\angle GYQ$

Therefore $\bigtriangleup PXG \sim \bigtriangleup YQG$

$\frac{XG}{GQ}=\frac{PG}{YG}$

$GY=\frac{266 \times 255}{190} = 357$

$XY= GY - GX = 357 -190 = 167$

Thus the answer is $\boxed{167}$

We have: $XA.XP=XY^2-R^2_2=XY^2-YA^2$ and $YA.YP=XY^2-R^2_1=XY^2-XA^2$ .

Therefore, $XA.XP-XA^2=YA.YQ-YA^2$ or $XA.PA=YA.QA$ .

Since $\angle XAY= \angle QAP$ and $\frac{XA}{YA}=\frac{QA}{PA}$ , so $\Delta GQY \propto \Delta GXP$

Therefore, $\frac{GQ}{GX}=\frac{GY}{GP}$ and $GY=\frac{GQ.GP}{GX}=357$ , so $XY=GY-GX=167$

Note that since $\angle QAX = \angle PAY$ and $QX=XA, AY = YP$ , then $\angle XQA = \angle QAX = \angle YAP = \angle APY$ , and thus $\angle QXA = \angle AYP$ , or in other words, $\angle QXP = \angle QYP$ , making $QXYP$ a cyclic quadrilateral. Now, consider its circumcircle $\Gamma$ .

The power of $G$ to $\Gamma$ can be expressed in two ways, which is $GX\times GY$ and $GQ \times GP$ . We have: $GY = \frac{GQ \times GP}{GX} = \frac{255\times 266}{190} = 357$ and consequently $XY = GY - GX = 357 - 190 = 167.$

We prove that XYQP is cyclic. Since QXA and AYP are both isoceles, we find that angle XQA=angle XAQ=angle YAP=angle YPA. Therefore, it is cyclic, and the power of G with respect to that circle yields the answer.

We begin by noticing that $A$ lies on the radical axis of the two circles, and so

$\displaystyle QA\cdot AY=PA\cdot AX\Longrightarrow \frac{QA}{PA}=\frac{AY}{AX}.$

Also, $\measuredangle QAX=\measuredangle PAY$ . This, combined with the above ratio, shows that $\triangle QAX\sim \triangle PAY$ . Since similar triangles have equal angles, $\measuredangle YPA=\measuredangle XQA$ , but this means that quadrilateral $QPYX$ is cyclic. Since opposite angles in a cyclic quad are supplementary,

$\displaystyle \measuredangle GQX=\pi-\measuredangle PQX=\measuredangle XYP.$

Similarly,

$\displaystyle \measuredangle GXQ=\pi-\measuredangle QXY=\measuredangle QPY.$

It follows from these two pairs of equal angles that $\triangle GQX\sim\triangle GYP$ . Using the ratios of sides, and plugging in our known values,

$\displaystyle \frac{GP}{GX}=\frac{GY}{GQ}\Longrightarrow \frac{266}{190}=\frac{190+\overline{XY}}{255}.$

Solving gives $\overline{XY}=167$ .

Since $\angle XPY =\angle YAP = \angle XAQ = \angle XQY$ , we have $XQPY$ is an cyclic quadrilateral. It follows that $\angle GQX = \angle GYP$ . The two triangles $GQX$ and $GYP$ have a common angle $\angle YGP$ and $\angle GQX = \angle GYP$ , so these two triangles are similar. This leads to the ratio $\frac{GQ}{GY}=\frac{GX}{GP} \rightarrow GY=\frac{GP*GQ}{GX}=\frac{255*266}{190}=357$ . So $XY=GY-GX=357-190=167$