Three circle C₁, C₂ and C₃ with radii r₁, r₂ and r₃ (where r₁ < r₂ < r₃) are placed as shown in the given figure. What is the value of r₂?

Solution (By Examveda Team)

$$\eqalign{ & AB = \sqrt {4{r_1}.{r_2}} \cr & BC = \sqrt {4{r_2}.{r_3}} \cr & AC = \sqrt {{{\left( {{r_1} + 2{r_2} + {r_3}} \right)}^2} - {{\left( {{r_3} - {r_1}} \right)}^2}} \cr & AB + BC = AC \cr & \sqrt {4{r_1}.{r_2}} + \sqrt {4{r_2}.{r_3}} = \sqrt {{{\left( {{r_1} + 2{r_2} + {r_3}} \right)}^2} - {{\left( {{r_3} - {r_1}} \right)}^2}} \cr & {\text{Squaring both sides}} \cr & \Rightarrow 4{r_1}.{r_2} + 4{r_2}.{r_3} + 2\sqrt {16{r_1}.{r_2}^2.{r_3}} = {\left( {{r_1} + 2{r_2} + {r_3}} \right)^2} - {\left( {{r_3} - {r_1}} \right)^2} \cr & \Rightarrow 4{r_1}.{r_2} + 4{r_2}.{r_3} + 8{r_2}\sqrt {{r_1}.{r_3}} = {r_1}^2 + 4{r_2}^2 + {r_3}^2 + 4{r_1}.{r_2} + 4{r_2}.{r_3} + 2{r_1}.{r_3} - {r_3}^2 - {r_1}^2 + 2{r_1}.{r_3} \cr & \Rightarrow 8{r_2}\sqrt {{r_1}.{r_3}} = 4{r_2}^2 + 4{r_1}.{r_3} \cr & \Rightarrow 2{r_2}\sqrt {{r_1}.{r_3}} = {r_2}^2 + {r_1}.{r_3} \cr & {\text{Squaring both sides}} \cr & \Rightarrow 4{r_2}^2.{r_1}.{r_3} = {r_2}^4 + {r_1}^2.{r_3}^2 + 2{r_2}^2.{r_1}.{r_3} \cr & \Rightarrow {\left( {{r_2}^2 - {r_1}.{r_3}} \right)^2} = 0 \cr & \Rightarrow {r_2}^2 = {r_1}.{r_3} \cr & \Rightarrow {r_2} = \sqrt {{r_1}.{r_3}} \cr} $$