Lines in circles?!

Let A, B, C, and D be points on a circle such that AB = 11 and CD = 19.
Point P is on segment AB with AP = 6, and Q is on segment CD with CQ = 7.
The line through P and Q intersects the circle at X and Y. If PQ = 27, find XY

$\begin{array}{|rcll|} \hline \text{Let AB } &=& 11 \\ \text{Let AP } &=& 6 \\ \text{Let PB } &=& \text{ AB } - \text{ AP } \\ &=& 11 - 6 \\ &=& 5 \\\\ \text{Let CD } &=& 19 \\ \text{Let CQ } &=& 7 \\ \text{Let QD } &=& \text{ CD } - \text{ CQ } \\ &=& 19 - 7 \\ &=& 12 \\\\ \text{Let PQ } &=& 27 \\\\ \text{Let XP } &=& x \\ \text{Let YQ } &=& y \\ \hline \end{array}$

$\begin{array}{|lrcll|} \hline (1) & \text{AP } \cdot \text{PB } &=& x \cdot \text{PY } \quad & | \quad \text{PY } = y + \text{ PQ } \\ & 6 \cdot 5 &=& x \cdot ( y + \text{ PQ } ) \\ & 30 &=& x \cdot ( y + 27 ) \\ & x &=& \frac{30}{y + 27} \\\\ (2) & \text{CQ } \cdot \text{QD } &=& y \cdot \text{QX } \quad & | \quad \text{QX } = x + \text{ PQ } \\ & 7 \cdot 12 &=& y \cdot ( x + \text{ PQ } ) \\ & 84 &=& y \cdot ( x + 27 ) \quad & | \quad x = \frac{30}{y + 27} \\ & 84 &=& y \cdot ( \frac{30}{y + 27} + 27 ) \\ & 84 &=& y \cdot \left( \frac{30}{y + 27} + 27 \right) \\ & 84 &=& y \cdot \left( \frac{30 +27\cdot (y + 27 ) }{y + 27} \right) \\ & 84 &=& y \cdot \left( \frac{30 +27y +27^2 }{y + 27} \right) \\ & 84 &=& y \cdot \left( \frac{759 +27y}{y + 27} \right) \\ & 84\cdot(y + 27) &=& y \cdot ( 759 +27y ) \\ & 84y + 84\cdot 27 &=& 759y +27y^2 \\ & 27y^2 +675y - 84\cdot 27 &=& 0 \quad & | \quad : 27 \\ & y^2 + 25y - 84 &=& 0 \\\\ & y_{1,2} &=& \frac{-25 \pm \sqrt{25^2 -4 \cdot(-84) } } {2} \\ & y_{1,2} &=& \frac{-25 \pm 31 } {2} \quad & | \quad y > 0!\\ & y &=& \frac{-25 + 31 } {2} \\ & \mathbf{y} &\mathbf{=}& \mathbf{3} \\\\ & x &=& \frac{30}{y + 27} \quad & | \quad y = 3 \\ & x &=& \frac{30}{3 + 27} \\ & x &=& \frac{30}{30} \\ & \mathbf{x} &\mathbf{=}& \mathbf{1} \\\\ & \text{XY } &=& x + \text{ PQ } + y \\ & \text{XY } &=& 1 + 27 + 3 \\\\ & \mathbf{ \text{XY } } & \mathbf{=} & \mathbf{31} \\ \hline \end{array}$