Let $B$ be the centre of the circle $x^2+y^2-2x+4y+1=0.$ Let the tangents at two points $P$ and $Q$ on the circle intersect at the point $A(3, 1).$ Then $8\left(\frac{\mathrm{area} \Delta \mathrm{APQ}}{\mathrm{area} \Delta \mathrm{BPQ}}\right)$ is equal to                 .

Given, centre of circle $x^2+y^2-2x+4y+1=0 .....\left(i\right)$

Points $B \left(1,-2\right) \& A\left(3,1\right)$

For point $P, y=-2$

Putting value of $y$ in equation $\left(i\right)$

$x^2+{\left(-2\right)}^2-2x+4\times \left(-2\right)+1=0$

$x^2-2x-3=0$

$\left(x-3\right)\left(x+1\right)=0$

$x=-1$ (not possible)

$x=3$

Point $P\left(3,-2\right)$

$AP=\sqrt{{\left(3-3\right)}^2+{\left(-2-1\right)}^2}$

$AP=3=AQ$

$r=\sqrt{{\left(1\right)}^2+{\left(-2\right)}^2-1}$

$r=\sqrt{1+4-1}$

$r=2$
$\tan \theta =\frac{3}{2}$

$\frac{Area\left(\Delta APQ\right)}{Area\left(\Delta BPQ\right)}=\frac{AR}{RB}$

$=\frac{3\sin \theta }{2\cos \theta }=\frac{9}{4}$

$8\left(\frac{Area\left(\Delta APQ\right)}{Area\left(\Delta BPQ\right)}\right)=18$