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If \mathrm{r}_1  and \mathrm{r}_2  are radii of the circles whose centres are at a distance 1 unit from the origin and touches the lines \mathrm{y}=0  and \mathrm{y}=\sqrt{3}(\mathrm{x}+1)  then \frac{\mathrm{r}_1}{\mathrm{r}_2}  is equal to
 

Option: 1

\frac{3}{2}


Option: 2

1


Option: 3

\frac{3}{4}


Option: 4

\frac{9}{16}


Answers (1)

best_answer

Centre of one circle will be at point A which is

intersection point of line \mathrm{y=\frac{1}{\sqrt{3}}(x+1)}  and  \mathrm{x^2+y^2=1}

solving we get A \equiv\left(\frac{1}{2}, \frac{\sqrt{3}}{2}\right) \Rightarrow r_1=\frac{\sqrt{3}}{2}

Centre of 2^{\text {nd }}  circle will be at point B which is intersection point of line \mathrm{y=-\sqrt{3}(x+1) \text{ and }x^2+y^2=1}

solving we get B \mathrm{ \equiv\left(-\frac{1}{2},-\frac{\sqrt{3}}{2}\right) \Rightarrow r_2=\frac{\sqrt{3}}{2} ~ therefore ~ \frac{r_1}{r_2}=1}

Posted by

Deependra Verma

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