Circle And Chords

Let $|PA| = a$ and $|PC| = b$ . Then by the Intersecting Secant theorem we have that

$|PA||PB| = |PC||PD| \Longrightarrow a(a + 8) = b(b + 12)$ , (i).

Next, since $ABCD$ is a cyclic quadrilateral we know that $\angle ABD + \angle ACD = 180^{\circ}$ . But as $\angle ACD + \angle ACP = 180^{\circ}$ as well we have that $\angle ABD = \angle ACP$ . Similarly $\angle CDB = \angle CAP$ . Thus triangles $\Delta PCA$ and $\Delta PBD$ are similar, in which case

$\dfrac{|PA|}{|AC|} = \dfrac{|PD|}{|DB|} \Longrightarrow \dfrac{a}{3} = \dfrac{b + 12}{5} \Longrightarrow b + 12 = \dfrac{5}{3}a$ .

Combining this with the equation (i) gives us that

$a^{2} + 8a = \left(\dfrac{5}{3}a - 12\right)\left(\dfrac{5}{3}a\right) \Longrightarrow 9a^{2} + 72a = (5a - 36)(5a)$

$\Longrightarrow 9a + 72 = 5(5a - 36) \Longrightarrow 9a + 72 = 25a - 180$

$\Longrightarrow 16a = 252 \Longrightarrow a = \dfrac{252}{16} = \dfrac{63}{4} = \boxed{15.75}$ .