Question

In the adjoining figure $\angle ACE$ is a right angle there are three circles which just touch each other and AC and EC are the tangents to all the three circles. What is the ratio of radii of the largest circle to that of the smallest circle ?

• A. $17:12\sqrt{2}$
• B. $1:(17-12\sqrt{2})$
• C. $12:17\sqrt{2}$
• D. $Noneoftheabove$

Answer 1

The correct option is B $1:(17-12\sqrt{2})$


In $\Delta CIE$,
$C{I}^2=C{E}^2+E{I}^2$
But CE = EI
$\therefore C{I}^2=2E{I}^{2}CI=\sqrt{2}EI=\sqrt{2}{r}_{3}CI=\sqrt{2}{r}_1+r_1+2r_2+r_3=(\sqrt{2}+1)r_1+2r_2+r_3(\sqrt{2}+1)r_1+2r_2+r_3=\sqrt{2}{r}_3(\sqrt{2}+1)r_1+2r_2=(\sqrt{2}-1)r_3\cdots \left(i\right)Consider,\Delta CHD,(\sqrt{2}+1)r_1=(\sqrt{2}-1)r_2{r}_2=\left(\frac{\sqrt{2}+1}{\sqrt{2}-1}\right)r_1$
Putting the value of $r_2$ in i), we get
$(\sqrt{2}+1)r_1+2\left\{\frac{\sqrt{2}+1}{\sqrt{2}-1}\right\}{r}_1=(\sqrt{2}-1)r_3(3+2\sqrt{2})r_1=(3-2\sqrt{2})r_3\frac{r_3}{r_1}=\frac{3+2\sqrt{2}}{3-2\sqrt{2}}=(3+2\sqrt{2}{)}^2=17+12\sqrt{2}\frac{r_3}{r_1}=\left(\frac{1}{17-12\sqrt{2}}\right)$