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 $$8 \cdot \frac{\text{area } \triangle APQ}{\text{area } \triangle BPQ}$$ is equal to _________.

We have the circle

$$x^{2}+y^{2}-2x+4y+1=0.$$

The general form of a circle is written as

$$x^{2}+y^{2}+2gx+2fy+c=0,$$

so by comparing we see

$$2g=-2 \;\;\Longrightarrow\;\; g=-1,$$

$$2f=4 \;\;\Longrightarrow\;\; f=2,$$

$$c=1.$$

Hence the centre of the circle is

$$B(-g,\,-f)=(1,\,-2).$$

The two tangents drawn from the external point

$$A(3,\,1)$$

touch the circle at the points $$P$$ and $$Q$$. The line $$PQ$$ is therefore the chord of contact of $$A$$ with respect to the given circle.

The formula for the chord of contact from an external point $$(x_{1},y_{1})$$ to the circle $$x^{2}+y^{2}+2gx+2fy+c=0$$ is obtained by replacing $$x^{2}$$ by $$xx_{1}$$ and $$y^{2}$$ by $$yy_{1}$$ in the equation of the circle. Symbolically,

$$T=0:\; xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0.$$

Here $$(x_{1},y_{1})=(3,\,1)$$. Substituting $$g=-1,\;f=2,\;c=1$$ we get

$$xx_{1}+yy_{1}+g(x+x_{1})+f(y+y_{1})+c=0,$$

$$x\cdot3+y\cdot1+(-1)(x+3)+2(y+1)+1=0.$$

Simplifying term by term:

$$3x+y-x-3+2y+2+1=0,$$

$$(3x-x)=2x,\quad (y+2y)=3y,\quad (-3+2+1)=0,$$

so the chord of contact is

$$2x+3y=0.$$

Thus the straight line joining $$P$$ and $$Q$$ is

$$PQ:\;2x+3y=0.$$

Both triangles $$\triangle APQ$$ and $$\triangle BPQ$$ have the same base $$PQ$$. Therefore the ratio of their areas equals the ratio of the perpendicular distances of the vertices $$A$$ and $$B$$ from the line $$PQ$$. That is,

$$\frac{\text{area }\triangle APQ}{\text{area }\triangle BPQ} =\frac{\text{distance of }A\text{ from }PQ} {\text{distance of }B\text{ from }PQ}.$$

The distance of a point $$(x_{0},y_{0})$$ from the line $$2x+3y=0$$ is given by the point-to-line distance formula:

$$d=\frac{|2x_{0}+3y_{0}|}{\sqrt{2^{2}+3^{2}}} =\frac{|2x_{0}+3y_{0}|}{\sqrt{13}}.$$

Distance of $$A(3,1)$$ from $$PQ$$

$$d_{A}=\frac{|2\cdot3+3\cdot1|}{\sqrt{13}} =\frac{|6+3|}{\sqrt{13}} =\frac{9}{\sqrt{13}}.$$

Distance of $$B(1,-2)$$ from $$PQ$$

$$d_{B}=\frac{|2\cdot1+3\cdot(-2)|}{\sqrt{13}} =\frac{|2-6|}{\sqrt{13}} =\frac{4}{\sqrt{13}}.$$

Hence

$$\frac{\text{area }\triangle APQ}{\text{area }\triangle BPQ} =\frac{d_{A}}{d_{B}} =\frac{9/\sqrt{13}}{4/\sqrt{13}} =\frac{9}{4}.$$

We are asked to find

$$8\cdot\frac{\text{area }\triangle APQ}{\text{area }\triangle BPQ} =8\cdot\frac{9}{4} =\frac{8\times9}{4} =2\times9 =18.$$

So, the answer is $$18$$.