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Question: Answered & Verified by Expert
Let $n \geq 3$ and let $C_{1}, C_{2}, \ldots . C_{n}$, be circles with radii $r_{1}, r_{2}, \ldots, r_{n}$, respectively. Assume that $C_{i}$ and $C_{i+1}$ touch externally for $1 \leq i \leq n-1$. It is also given that the $x$-axis and the line $y=2 \sqrt{2} x+10$ are tangential to each of the circles. Then $r_{1}, r_{2}, \ldots \ldots, r_{n}$ are in
MathematicsCircleJEE Main
Options:
  • A An arithmetic progression with common difference $3+\sqrt{2}$
  • B A geometric progression with common ratio $3+\sqrt{2}$
  • C An arithetic progression with common diffrence $2+\sqrt{3}$
  • D Ageometric progression with common ratio $2+\sqrt{3}$
Solution:
2033 Upvotes Verified Answer
The correct answer is: Ageometric progression with common ratio $2+\sqrt{3}$


$\operatorname{Sin} \theta=\frac{1}{\sqrt{3}}=\frac{r_{1}}{\mathrm{AP}_{1}}=\frac{r_{2}}{\mathrm{AP}_{2}}$
$\mathrm{AP}_{2}=\mathrm{AP}_{1}+\mathrm{r}_{1}+\mathrm{r}_{2}$
$\mathrm{r}_{2} \sqrt{3}=\mathrm{r}_{1} \sqrt{3}+\mathrm{r}_{1}+\mathrm{r}_{2}$
$\mathrm{r}_{2}(\sqrt{3}-1)=\mathrm{r}_{1}(\sqrt{3}+1)$
$\frac{r_{2}}{r_{1}}=2+\sqrt{3}$

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