Question

Let $2x^2+y^2-3xy=0$ be the equation of pair of tangents drawn from the origin to a circle of radius $3$, with centre in $1^{st}$ quadrant. If A is one of the points of contact, find OA.

Answer 1

Given pair of tangents are $2x^2+y^2-3xy=0$ ........$\left(1\right)$

$2x^2-3xy+y^2=0$

$2x^2-2xy-xy+y^2=0$

$2x(x-y)-y(x-y)=0$

$(x-y)(2x-y)=0$

$\Rightarrow x-y=0$ or $2x-y=0$

Thus,$y=x$ and $y=2x$ are the two tangents of the circle at the points $B$ and $A$ respectively.

Let $m_1=1$ and $m_2=2$

Let $2\alpha$ be the angle between these two tangents.

Let $\angle AOB=2\alpha \Rightarrow \angle AOC=\alpha$

$\Rightarrow \tan 2\alpha =\frac{m_2-m_1}{1+m_1{m}_2}$

$=\frac{2-1}{1+2}=\frac{1}{3}$

$\Rightarrow \frac{2\tan \alpha }{1-{\tan }^2\alpha }=\frac{1}{3}$

$\Rightarrow 1-{\tan }^2\alpha =6\tan \alpha$

$\Rightarrow 1-6\tan \alpha -{\tan }^2\alpha =0$

$\tan \alpha =\frac{-6\pm \sqrt{36+4}}{2}=\frac{-6\pm 2\sqrt{10}}{2}=-3\pm \sqrt{10}$

as $\alpha$ lies between $0$ and $\frac{\pi }{4},\tan \alpha =\sqrt{10}-3$

the radius of the circle$=3$cm

$AC=3$cm

$\tan \alpha =\frac{AC}{OA}$

$\Rightarrow OA=\frac{AC}{\tan \alpha }=\frac{3}{\sqrt{10}-3}$

$\Rightarrow OA=\frac{3}{\sqrt{10}-3}\times \frac{\sqrt{10}+3}{\sqrt{10}+3}=\frac{3(\sqrt{10}+3)}{10-9}=\frac{3(\sqrt{10}+3)}{1}=3(\sqrt{10}+3)$