[Geometry]
Problem: Circle $O$ has radius 6. Point $P$ lies outside circle $O$, and the shortest distance from $P$ to circle $O$ is 4. Chord $\overline{A B}$ is parallel to $\overleftrightarrow{O P}$, and the distance between $\overline{A B}$ and $\overleftrightarrow{O P}$ is 2 . Compute $P A^{2}+P B^{2}$.

Solution: Extend $\overline{A B}$ to point $Q$ such that $\overline{P Q} \perp \overline{A Q}$ as shown, and let $M$ be the midpoint of $\overline{A B}$. (The problem does not specify whether $A$ or $B$ is nearer $P$, but $B$ can be assumed to be nearer $P$ without loss of generality.)

<img_3454>

Then $O P=10, P Q=O M=2$, and $O B=6$. Thus $M B=\sqrt{6^{2}-2^{2}}=4 \sqrt{2}$. Because $Q M=O P=10$, it follows that $Q B=10-4 \sqrt{2}$ and $Q A=10+4 \sqrt{2}$. So

$$
\begin{aligned}
P A^{2}+P B^{2} & =\left(Q A^{2}+Q P^{2}\right)+\left(Q B^{2}+Q P^{2}\right) \\
& =(10+4 \sqrt{2})^{2}+2^{2}+(10-4 \sqrt{2})^{2}+2^{2} \\
& =\mathbf{2 7 2}
\end{aligned}
$$

Final answer: 272