Three circle with radius $r_1$, $r_2$ and $r_3$ where ($r_1$<$r_2$<$r_3$) touch each other externally . If they have a common tangent then the value of $\sqrt{\frac{r_1}{r_2}}+\sqrt{\frac{r_1}{r_3}}$=__________
My approach is illustrated in the diagram
How will I proceed from here
Length of common tangent (CT) of two touching circlesis $L=2\sqrt{r_1r_2}$. CT of 1 and 3 is $L_{13}=2\sqrt{r_1r_3}$, CT of 1 and 2 is $L_{12}= \sqrt{r_1r_2}$, CT of 2 and 3 is $L_{23}=2\sqrt{r_2r_3}.$
Next, $$L_{23}=L_{12}+L_{13}\implies 2 \sqrt{r_2r_3}=2\sqrt{r_1r_3}+2 \sqrt{r_1r_2}$$ Hence we get $$\sqrt{\frac{r_1}{r_2}}+\sqrt{\frac{r_1}{r_3}}=1$$
Hint:
In the diagram, you will notice by Pythagoras that
$(r_3 + r_2)^2 = (r_3 -r_2)^2 + (t_{23})^2$ where $t_{23}$ is the length of the tangent.
So $t_{23} = 2 \sqrt{r_3 r_2}$
Similarly $t_{12} = 2 \sqrt{r_1 r_2}, t_{13} = 2 \sqrt{r_1 r_3}$
Now $t_{23} = t_{12} + t_{13}$ and that leads you to the answer.
