Question

Le $n\ge 3$ and let $C_1,C_2,....C_n$, be circles with radii $r_1,r_2,.....r_n$, respectively. Assume that $C_i$ and $C_{i+1}$ touch externally for $1\le i\le n-1$. 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 $r_1,r_2,.....,r_n$ are in.

• A. An arithmetic progression with common difference $3+\sqrt{2}$
• B. A geometric progression with common ratio $3+\sqrt{2}$
• C. An arithmetic progression with common difference $2+\sqrt{3}$
• D. A geometric progression with common ratio $2+\sqrt{3}$

Answer 1

Answer: (D)

A geometric progression with common ratio $2+\sqrt{3}$

Let us solve the problem with $n=2$ and we will get a generalised result

Let the angle between $y=2\sqrt{2}x+10$ and $x$ axis be $2\theta$

Angle between a line and $x$ axis that is $\tan \theta$ is equal to the slope of the given line

$\Rightarrow \tan 2\theta =2\sqrt{2}\Rightarrow \cos 2\theta =\frac{1}{3}1-2{\sin }^2\theta =\frac{1}{3}\Rightarrow \sin \theta =\frac{1}{\sqrt{3}}$

From the figure

$\sin \theta =\frac{r_1}{{AP}_1}=\frac{r_2}{{AP}_2}=\frac{1}{\sqrt{3}}.....\left(i\right){AP}_2={AP}_1+P_1{P}_2{AP}_2={AP}_1+r_1+r_2\sqrt{3}{r}_2=\sqrt{3}{r}_1+r_1+r_2{r}_2(\sqrt{3}-1)=r_1(\sqrt{3}+1)\frac{r_2}{r_1}=\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}=\frac{(\sqrt{3}+1)}{(\sqrt{3}-1)}\times \frac{(\sqrt{3}+1)}{(\sqrt{3}+1)}=2+\sqrt{3}$

Clealry the radius are in geometric progression and the result will also be followed for the proceeding circles.

Hence, option $D$ is correct