Circle 5 Question 16
16. Let $2 x^{2}+y^{2}-3 x y=0$ be the equation of a pair of tangents drawn from the origin $O$ to a circle of radius 3 with centre in the first quadrant. If $A$ is one of the points of contact, find the length of $O A$.
$(2001,5$ M)
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Answer:
Correct Answer: 16. $3(3+\sqrt{10})$
Solution:
$ 2 x^{2}+y^{2}-3 x y=0 $
[given]
$ \begin{aligned} \Rightarrow & & 2 x^{2}-2 x y-x y+y^{2} & =0 \\ & \Rightarrow & 2 x(x-y)-y(x-y) & =0 \\ & \Rightarrow & (2 x-y)(x-y) & =0 \\ & \Rightarrow & y & =2 x, y=x \end{aligned} $
are the equations of straight lines passing through origin.
Now, let the angle between the lines be $2 \theta$ and the line $y=x$ makes angle of $45^{\circ}$ with $X$-axis.
Therefore, $\tan \left(45^{\circ}+2 \theta\right)=2$ (slope of the line $\left.y=2 x\right)$
$ \begin{aligned} & \xrightarrow{x _1} \\ & \Rightarrow \quad \frac{\tan 45^{\circ}+\tan 2 \theta}{1-\tan 45^{\circ} \times \tan 2 \theta}=2 \Rightarrow \frac{1+\tan 2 \theta}{1-\tan 2 \theta}=2 \\ & \Rightarrow \frac{(1+\tan 2 \theta)-(1-\tan 2 \theta)}{(1+\tan 2 \theta)+(1-\tan 2 \theta)}=\frac{2-1}{(2+1)}=\frac{1}{3} \\ & \Rightarrow \quad \frac{2 \tan 2 \theta}{2}=\frac{1}{3} \Rightarrow \tan 2 \theta=\frac{1}{3} \\ & \Rightarrow \quad \frac{2 \tan \theta}{1-\tan ^{2} \theta}=\frac{1}{3} \\ & \Rightarrow \quad(2 \tan \theta) \cdot 3=1-\tan ^{2} \theta \\ & \Rightarrow \tan ^{2} \theta+6 \tan \theta-1=0 \\ & \Rightarrow \quad \tan \theta=\frac{-6 \pm \sqrt{36+4 \times 1 \times 1}}{2}=\frac{-6 \pm \sqrt{40}}{2} \\ & \Rightarrow \quad \tan \theta=-3 \pm \sqrt{10} \\ & \Rightarrow \quad \tan \theta=-3+\sqrt{10} \quad \because 0<\theta<\frac{\pi}{4} \\ & \tan \theta=\frac{3}{O A} \\ & \therefore \quad O A=\frac{3}{\tan \theta}=\frac{3}{(-3+\sqrt{10})}=\frac{3(3+\sqrt{10})}{(-3+\sqrt{10})(3+\sqrt{10})} \\ & =\frac{3(3+\sqrt{10})}{(10-9)}=3(3+\sqrt{10}) \end{aligned} $
