The tangent to the circumcircle of triangle $WXY$ at $X$ is drawn, and the line through $W$ that is parallel to this tangent intersects $\overline{XY}$ at $Z.$ If $XY = 14$ and $WX = 6,$ find $YZ.$ [asy] unitsize(2 cm); pair A, B, C, D; A = dir(110); B = dir(210); C = dir(330); D = extension(B, C, A, A + rotate(90)*(B)); draw(Circle((0,0),1)); draw(A--B--C--cycle); draw((B + rotate(90)*(B))--(B - rotate(90)*(B))); draw(A--(A + 2.2*(rotate(90)*(B)))); label("$W$", A, N); label("$X$", B, SW); label("$Y$", C, SE); label("$Z$", D, SW); [/asy]
Let O be the center of the circumcircle of triangle WXY, and let M be the midpoint of XY. Since O is the circumcenter, we have OW = OX = OY, and so triangle WOX is isosceles. Therefore, angle WOX = angle WCO, where C is the midpoint of WX.
Since the tangent to the circumcircle at X is perpendicular to OX, we have angle OXM = 90 degrees. Since M is the midpoint of XY, we have MX = MY = 7, and so angle OMY = arctan(7/OW). Thus, angle OXY = angle WOX - angle WOC = angle OMY - angle WOC = arctan(7/OW) - angle WOC.
Since the line through W that is parallel to the tangent intersects XY at Z, we have angle YZW = angle WOC. Therefore, angle YZX = angle OXY + angle YZW = arctan(7/OW). Using the fact that WX = 6, we can use the Pythagorean theorem to find that OM = sqrt(7^2 + 3^2) = sqrt(58), and so OW = 2OM = 2sqrt(58).
Thus, angle YZX = arctan(7/(2sqrt(58))), and we can use the tangent addition formula to find that tan(angle YZX) = (2sqrt(58))/7. Since YZ = XY - XZ and XZ = WX tan(angle YZX), we have YZ = 14 - 6(2sqrt(58))/7 = (98 - 12sqrt(58))/7.
Therefore, YZ = (98 - 12sqrt(58))/7.
