Sum of all the radii of the circles touching the coordinate axes and the line 3x + 4y = 12, is
(A) 1                      (B) 2                       (C) 12                     (D) $\frac{9}{2}$

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Please find below the solution to the asked query:

$$\begin{gathered} \mathrm{Let} \mathrm{O}\hspace{0.17em}\mathrm{be} \mathrm{origin} \mathrm{and} \mathrm{line} \mathrm{meets} \mathrm{x} \mathrm{and} \mathrm{y} \mathrm{axis} \mathrm{at} \mathrm{A} \mathrm{and} \mathrm{B} \mathrm{respectively} \\ \mathrm{There} \mathrm{are} 4 \mathrm{circles} \mathrm{which} \mathrm{touch} \mathrm{triangle} \mathrm{OAB}, \mathrm{one} \mathrm{incircle} \mathrm{and} 3 \mathrm{excircles}. \\ \mathrm{Now} \\ 3\mathrm{x} + 4\mathrm{y} = 12 \\ \Rightarrow \frac{3\mathrm{x}}{12}+\frac{4\mathrm{y}}{12}=1 \\ \Rightarrow \frac{\mathrm{x}}{4}+\frac{\mathrm{y}}{3}=1 \\ \mathrm{Hence} \mathrm{by} \mathrm{intercept} \mathrm{form}, \mathrm{co}-\mathrm{ordinates} \mathrm{of} \mathrm{A}\hspace{0.17em}\mathrm{is} \left(4,0\right) \mathrm{and} \mathrm{B}\left(0,3\right) \\ \mathrm{Hence} \\ \mathrm{OA}=4=\mathrm{a} \\ \mathrm{OB}=3=\mathrm{b} \\ \mathrm{AB}=\sqrt{{\left(4-0\right)}^2+{\left(0-3\right)}^2}=\sqrt{25}=5=\mathrm{c} \\ \mathrm{Area}=△=\frac{1}{2}\times \mathrm{OA}\times \mathrm{OB}=\frac{1}{2}\times 4\times 3=6 \\ \mathrm{Semi}-\mathrm{perimeter}=\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}=\frac{4+3+5}{2}=6 \\ \mathrm{Now} \\ \mathrm{Inradius}=\mathrm{r}=\frac{△}{\mathrm{s}}=\frac{6}{6}=1 \\ \mathrm{Exradii} \mathrm{are} \mathrm{given} \mathrm{by} \\ {\mathrm{r}}_1=\frac{△}{\mathrm{s}-\mathrm{a}}=\frac{6}{6-4}=3 \\ {\mathrm{r}}_2=\frac{△}{\mathrm{s}-\mathrm{b}}=\frac{6}{6-3}=2 \\ {\mathrm{r}}_3=\frac{△}{\mathrm{s}-\mathrm{c}}=\frac{6}{6-5}=6 \\ \mathrm{Required} \mathrm{sum}=1+3+2+6=12 \left[\mathrm{Answer}\right] \end{gathered}$$

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