10. Suppose $\omega$ is a circle centered at $O$ with radius 8 . Let $A C$ and $B D$ be perpendicular chords of $\omega$. Let $P$ be a point inside quadrilateral $A B C D$ such that the circumcircles of triangles $A B P$ and $C D P$ are tangent, and the circumcircles of triangles $A D P$ and $B C P$ are tangent. If $A C=2 \sqrt{61}$ and $B D=6 \sqrt{7}$, then $O P$ can be expressed as $\sqrt{a}-\sqrt{b}$ for positive integers $a$ and $b$. Compute $100 a+b$.
Answer:
103360

Solution: Let $X=A C \cap B D, Q=A B \cap C D$ and $R=B C \cap A D$. Since $Q A \cdot Q B=Q C \cdot Q D, Q$ is on the radical axis of $(A B P)$ and $(C D P)$, so $Q$ lies on the common tangent at $P$. Thus, $Q P^{2}=Q A \cdot Q B$. Similarly, $R A \cdot R C=R P^{2}$. Let $M$ be the Miquel point of quadrilateral $A B C D$ : in particular, $M=O X \cap Q R$ is the foot from $O$ to $Q R$. By properties of the Miquel point, $A B M R$ and $A C M Q$ are cyclic. Thus,
$$
\begin{aligned}
Q P^{2} & =Q A \cdot Q B \\
R P^{2} & =R A \cdot R C \\
Q P^{2}+R P^{2} & =Q M \cdot Q R+R M \cdot R Q=(Q R+R M) Q R=Q R^{2} .
\end{aligned}
$$

As a result, $\angle Q P R=90^{\circ}$.
Now, let $P^{\prime}$ the inverse of $P$ with respect to $\omega$. Note that by properties of inversion, $\left(A B P^{\prime}\right)$ and $\left(C D P^{\prime}\right)$ are tangent, and $\left(A C P^{\prime}\right)$ and $\left(B D P^{\prime}\right)$ are also tangent.
But now,
$$
\begin{aligned}
Q P^{2}=Q P^{\prime 2}=Q A \cdot Q B \\
R P^{2}=R P^{\prime 2}=R A \cdot R C \\
Q P^{2}+R P^{2}=Q P^{\prime 2}+R P^{\prime 2}=Q R^{2} .
\end{aligned}
$$

Thus, $P Q P^{\prime} R$ is a cyclic kite, so $P$ and $P^{\prime}$ are reflections of each other across $Q R$. In particular, since $O, P, P^{\prime}$ are collinear, then $M$ lies on line $O P P^{\prime}$.
We can now compute $O P$ by using the fact that $O P+\frac{r^{2}}{O P}=2 O M=\frac{2 r^{2}}{O X}$, where $r=8$. Since $O X$ can be computed to equal 2 quite easily, then $O P+\frac{64}{O P}=64$, or $O P^{2}-64 O P+64=0$. Solving this yields $O P=32 \pm 8 \sqrt{15}$, and because $P$ is inside the circle, $O P=32-8 \sqrt{15}=\sqrt{1024}-\sqrt{960}$.