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 $8\cdot \left(\frac{\text{area}△APQ}{\text{area}△BPQ}\right)$ is equal to

Answer 1

Let $L=$ length of tangent from $A$ to the circle and $R=$ radius of circle
$\angle PAB=\angle BPQ=\theta$
$\Rightarrow \text{area of}△PAQ=2\cdot \frac{1}{2}\cdot L\sin \theta \cdot Lcos\theta =L^2\sin \theta \cos \theta$
$\text{area of}△PBQ=2\cdot \frac{1}{2}\cdot R\sin \theta \cdot R\cos \theta =R^2\sin \theta \cos \theta$
Hence, $\frac{\text{area of}△APQ}{\text{area of}△BPQ}=\frac{L^2}{R^2}$
Now, $L=\sqrt{S_1}=\sqrt{3^2+1^2-2\times 3+4\times 1+1}=3$
and $R=2$
$\Rightarrow 8\times \left(\frac{\text{area of}△APQ}{\text{area of}△BPQ}\right)=8\times {\left(\frac{3}{2}\right)}^2=18$