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  • Two fixed circles with radii and <img alt="\mathrm{r_2\left(r_1>r_2\right)}" src="https://entr

Two fixed circles with radii \mathrm{r_1} and \mathrm{r_2\left(r_1>r_2\right)} respectively touch each other externally. Then the locus of the point of intersection of their direct common tangents is
 

Option: 1

 a circle
 


Option: 2

 a straight line


Option: 3

 a parabola


Option: 4

 none of these


Answers (1)

best_answer

 Let \mathrm{A \equiv(0,0)} and \mathrm{B \equiv\left(r_1+r_2, 0\right)} be the centres of the two given fixed circles.

Let \mathrm{C \equiv(\alpha, 0)} be the point of intersection of direct common tangents.

Now, \mathrm{\frac{r_2}{r_1}=\frac{\alpha-\left(r_1+r_2\right)}{\alpha} \Rightarrow r_2 \alpha=r_1 \alpha-r_1^2-r_1 r_2 \Rightarrow \alpha=\frac{r_1^2+r_1 r_2}{r_1-r_2} }

\therefore  Locus of C is \mathrm{x=\frac{r_1^2+r_1 r_2}{r_1-r_2}=a} constant, which is a straight line.

Posted by

Shailly goel

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