Let four circles having radii $r_1=5 \mathrm{units}, r_2=5 \mathrm{units}, r_3=8 \mathrm{units}$ and $r_4 \mathrm{units}\left(r_4<5\right)$ are mutually touching each other externally, then the value of $\frac{2}{r_4}$ is equal to

Let, the centres of circles with radii $r_1,r_2,r_3,r_4$ are $C_1,C_2,C_3$ and $C_4$ respectively.

The foot of $C_3$ on $C_1{C}_2$ is $E,$ then $C_{3}E=\sqrt{13^2-5^2}=12$, $C_{4}E=12-8-r_{{\mathrm{}}_4}=4-r_4.$
Now, ${\left(C_4{C}_2\right)}^2={\left(C_{4}E\right)}^2+5^2\Rightarrow {\left(5+r_4\right)}^2={\left(4-r_4\right)}^2+5^2$

$\Rightarrow r_4=\frac{8}{9}\Rightarrow \frac{2}{r_4}=\frac{9}{4}=2.25$