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  • If a circle of radius 3 units is touching the lines <img alt="\mathrm{\sqrt{3} y^2-4 x y+\sqrt{3} x^2=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%5Csqrt%7B3%7D%20y%5E

If a circle of radius 3 units is touching the lines \mathrm{\sqrt{3} y^2-4 x y+\sqrt{3} x^2=0}  in the first quadrant then length of chord of contact to this circle is

Option: 1

\frac{\sqrt{3}+1}{2}


Option: 2

\frac{\sqrt{3}+1}{\sqrt{2}}


Option: 3

3\left(\frac{\sqrt{3}+1}{\sqrt{2}}\right)


Option: 4

3\left(\frac{\sqrt{3}+1}{{2}}\right)


Answers (1)

best_answer

Given equation of lines

\mathrm{\begin{aligned} & \sqrt{3} y^2-4 x y+\sqrt{3} x^2=0 \\ & \sqrt{3} y^2-3 x y-x y+\sqrt{3} x^2=0 \\ & \Rightarrow(\sqrt{3} y-x)(y-\sqrt{3} x)=0 \Rightarrow y=\frac{x}{\sqrt{3}} \\ & y=\sqrt{3} x \\ & \angle A P O=75^{\circ} \end{aligned}}

Length of chord of contact AB 

=\mathrm{2 \cdot 3 \sin 75^{\circ}=6\left(\sin 45^{\circ} \cos 30^{\circ}+\sin 30^{\circ} \cos 45^{\circ}\right)}

\mathrm{=6\left(\frac{1}{\sqrt{2}} \cdot \frac{\sqrt{3}}{2}+\frac{1}{2} \cdot \frac{1}{\sqrt{2}}\right)=\frac{6(\sqrt{3}+1)}{2 \sqrt{2}}=\frac{3(\sqrt{3}+1)}{\sqrt{2}}}

 

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Gaurav

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