Equation of circle is $x^2+y^2-2 x+4 y+3=0$
$\therefore$ Length of $P A=\sqrt{S_1}$
$\begin{aligned} & =\sqrt{(6)^2+(-5)^2-2(6)+4(-5)+3} \\ & =\sqrt{56+25-12-20+3} \\ & =\sqrt{32}=4 \sqrt{2}\end{aligned}$
and radius, $O A=\sqrt{(1)^2+(-2)^2-3}=\sqrt{1+4-3}=\sqrt{2}$
$\therefore$ In $\triangle A O P$,
$\begin{gathered}\tan \frac{\theta}{2}=\frac{O A}{A P}=\frac{\sqrt{2}}{4 \sqrt{2}}=\frac{1}{4} \\ \therefore \tan \theta=\frac{2 \tan \theta / 2}{1-\tan ^2 \theta / 2}=\frac{2 \times \frac{1}{4}}{1-\left(\frac{1}{4}\right)^2}=\frac{\frac{1}{15}}{\frac{15}{16}}=\frac{8}{15}\end{gathered}$