Fourteen, in Figure 6, the radii of the two circles are 8 and 6, the distance between the two centers is 12, a line passing through one of the intersection points of the two circles intercepts equal chords $\mathrm{QP}$ and $\mathrm{PR}$ on the two circles, find the square of the length of $QP$.
Solve as shown in Figure 7, construct $\mathbf{O}_{1} M \perp Q \mathbf{Q}$ at $M, \mathbf{O}_{2} \mathbf{N} \perp$ $Q R$ at. Given $Q P=$
$\mathrm{PR}$, then $\mathrm{QM}=\mathrm{NR}=\frac{1}{2} \mathrm{Q} P$.
Let $Q \mathrm{M}=\mathrm{x}$, then
$$
\begin{array}{c}
\mathrm{O}_{1} \mathrm{M}=\sqrt{ } \overline{\mathrm{Q}^{2}-\mathrm{O}_{1} \mathrm{M}^{2}} \\
=\sqrt{8^{2}-\overline{x^{2}}},
\end{array}
$$
$$
\mathbf{O}_{2} \mathbf{N}=\sqrt{ } \mathrm{O}_{2}^{-} \mathrm{R}^{2}-\mathrm{NR}^{2}-\therefore 1 / \overline{6^{2}}-x^{2}
$$

Then $\mathrm{O}: \mathrm{X}=\mathrm{M}-\mathrm{O}_{2} \mathrm{~N}, \mathrm{O}_{2} \mathrm{~K}=\mathrm{MN}=2 \mathrm{x}$.
By the Pythagorean theorem, we get
$$
\mathrm{O}_{1} \mathrm{O}_{2}{ }^{2}=\mathrm{O}_{1} \mathrm{~K}^{2}+\mathrm{O}_{2} \mathrm{~K}^{2},
$$

which is $144=\left(\sqrt{ } 8^{2}-x^{2}-\sqrt{ } 6^{2}-x^{2}\right)^{2}+(2 x)^{2}$. Solving this, we get $4 \mathrm{x}^{2}=130$, which means $Q \mathrm{P}^{2}=130$.