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

Question

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 $8\cdot\dfrac{	ext{area}(	riangle APQ)}{	ext{area}(	riangle BPQ)}$ is equal to \_\_\_\_\_\_\_\_.

Answer: 18

Accepted Answer

Centre $B=(1,-2)$, $r=2$. $AB=\sqrt{(3-1)^2+(1+2)^2}=\sqrt{13}$; tangent length $AP=\sqrt{13-4}=3$.

Since both triangles share base $PQ$, the ratio of areas equals the ratio of heights from $A$ and $B$ to $PQ$, which (by similar right triangles in the tangent configuration) equals $\dfrac{AP\sin\theta}{BP\cos\theta}=\dfrac{3\sin\theta}{2\cos\theta}$ where $\tan\theta=\dfrac{AP}{BP}=\dfrac32$, giving $\sin\theta=\dfrac3{\sqrt{13}},\cos\theta=\dfrac2{\sqrt{13}}$:

$$\frac{\text{area}(\triangle APQ)}{\text{area}(\triangle BPQ)}=\frac{3\cdot\frac3{\sqrt{13}}}{2\cdot\frac2{\sqrt{13}}}=\frac94.$$

$$8\times\frac94=18.$$

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