Question

If the tangents drawn at the points $O(0,0)$ and $P(1+\sqrt{5},2)$ on the circle $x^2+y^2-2x-4y=0$ intersect at the point $Q$, then the area of the triangle $OPQ$ is equal to

Answer 1

$\tan 2\theta =2\Rightarrow \frac{2\tan \theta }{1-{\tan }^2\theta }=2$

$\tan \theta =\frac{\sqrt{5}-1}{2}\left(\text{as}\theta \text{is acute}\right)$

Area $=\frac{1}{2}{L}^2\sin 2\theta =\frac{1}{2}\cdot \frac{5}{{\tan }^2\theta }\cdot 2\sin \theta \cos \theta$

$=\frac{5\sin \theta \cos \theta }{{\sin }^2\theta }\cdot {\cos }^2\theta$

$=5\cot \theta \cdot {\cos }^2\theta$

$=5\cdot \frac{2}{\sqrt{5}-1}\cdot \frac{1}{1+{\left(\frac{\sqrt{5}-1}{2}\right)}^2}$

$=\frac{10}{\sqrt{5}-1}\cdot \frac{4}{4+6-2\sqrt{5}}$

$=\frac{40}{2\sqrt{5}(\sqrt{5}-1{)}^2}=\frac{4\sqrt{5}}{6-2\sqrt{5}}$

$=\frac{4\sqrt{5}(6+2\sqrt{5})}{16}$

$=\frac{\sqrt{5}(3+\sqrt{5})}{2}$