Let B be the centre of the circle x ^ 2 + y ^ 2 - 2x + 4y + 1 = 0 ....

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

Points $B\left(1, - 2\right) \text{ and } A\left(3, 1\right)$

For point $P, y = - 2$

Putting value of $y$ in equation \left(i\right)$</p><p>${x}^{2} + {\left(- 2\right)}^{2} - 2x + 4 \times \left(- 2\right) + 1 = 0$</p><p>${x}^{2} - 2x - 3 = 0$</p><p>\left(x - 3\right)\left(x + 1\right) = 0$

$x = - 1$ (not possible)

$x = 3$

Point $P\left(3, - 2\right)$

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

$AP = 3 = AQ$

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

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

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

$\dfrac{Area ( \delta APQ )}{Area ( \delta BPQ )} = \dfrac{AR}{RB}$

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

$8\left(\dfrac{Area ( \delta APQ )}{Area ( \delta BPQ )}\right) = 18$