13. $A B C D$ is a convex quadrilateral in which $A C$ and $B D$ meet at $P$. Given $P A=1$, $P B=2, P C=6$ and $P D=3$. Let $O$ be the circumcentre of $\triangle P B C$. If $O A$ is perpendicular to $A D$, find the circumradius of $\triangle P B C$.
(2 marks)
$A B C D$ is a convex quadrilateral, in which $A C$ and $B D$ intersect at $P$. Given $P A=1$, $P B=2$, $P C=6$, and $P D=3$. Let $O$ be the circumcenter of $\triangle P B C$. If $O A$ is perpendicular to $A D$, find the circumradius of $\triangle P B C$.
13. 3
13. Let \( O P \) meet \( A D \) at \( Q \). Note that we have \( O B = O C = O P \) and hence \( \angle O P B + \angle P C B = 90^\circ \). Since \( \angle O P B = \angle Q P D \) and \( \angle P C B = \angle P D Q \) (as \( P A \cdot P C = P B \cdot P D = 6 \), which implies \( A B C D \) is concyclic), we have \( \angle Q P D + \angle P D Q = 90^\circ \).
This implies \( O Q \) is perpendicular to \( A D \). Yet it is also given that \( O A \) is perpendicular to \( A D \). Hence \( O \) must lie on the straight line \( A P C \). As \( O \) is the circumcentre of \( \triangle P B C \), it follows that \( P C \) is a diameter of its circumcircle. Since \( P C = 6 \), the circumradius is 3.