Point $P$ is located inside a square $ABCD$ of side length $10$. Let $O_1$, $O_2$, $O_3$, $O_4$ be the circumcenters of $P AB$, $P BC$, $P CD$, and $P DA$, respectively. Given that $P A+P B +P C +P D = 23\sqrt2$ and the area of $O_1O_2O_3O_4$ is $50$, the second largest of the lengths $O_1O_2$, $O_2O_3$, $O_3O_4$, $O_4O_1$ can be written as $\sqrt{\frac{a}{b}}$, where $a$ and $b$ are relatively prime positive integers. Compute $100a + b$.

Let's think step by step and output the final answer within \boxed{}.
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