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  • A circle touches the lines <img alt="\mathrm{y=\frac{x}{\sqrt{3}}, y=\sqrt{3} x}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7By%3D%5Cfrac%7Bx%7D%7B%5Csqrt%7B3%7D%7D%2C%20y%3D%5

A circle touches the lines \mathrm{y=\frac{x}{\sqrt{3}}, y=\sqrt{3} x} the centre of this circle lies in the first quadrant then one possible equation of this circle is

Option: 1

\mathrm{x^{2}+y^{2}-2 x(\sqrt{3}+1)-2 y(\sqrt{3}+1)+8+4 \sqrt{3}=0}


Option: 2

\mathrm{x^{2}+y^{2}-2 x(\sqrt{3}+1)-2 y(\sqrt{3}+1)+5+4 \sqrt{3}=0}


Option: 3

\mathrm{x^{2}+y^{2}-2 x(\sqrt{3}+1)-2 y(\sqrt{3}+1)+7+4 \sqrt{3}=0}


Option: 4

\mathrm{x^{2}+y^{2}-2 x(\sqrt{3}+1)-2 y(\sqrt{3}+1)+6+4 \sqrt{3}=0}


Answers (1)

best_answer

Angle between lines is 60^{\circ}-30^{\circ}=30^{\circ}. Thus equation of their acute angle bisector is \mathrm{y} \tan \left(30^{\circ}+15^{\circ}\right). x i.e., \mathrm{y=x}.let


\begin{aligned} & \mathrm{C}=(\mathrm{h}, \mathrm{h}) \text { then } \frac{|\mathrm{h}-\mathrm{h} \sqrt{3}|}{2}=1 \Rightarrow(\sqrt{3}-1) \mathrm{h}=2 \\ & \Rightarrow \mathrm{h}=\frac{2}{(\sqrt{3}-1)}=\frac{2(\sqrt{3}+1)}{2}=(\sqrt{3}+1) \end{aligned}

thus equation of circle is
\begin{aligned} & (x-(\sqrt{3}+1))^{2}+(y-(\sqrt{3}+1))^{2}=1 \\ & \Rightarrow x^{2}+y^{2}-2 x(\sqrt{3}+1)-2 y(\sqrt{3}+1)+7+4 \sqrt{3}=0 \end{aligned}

Hence (C) is the correct answer.

Posted by

mansi

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