question_answer

Three circles of radii \[a,b,c(a<b<c)\] touch each other externally. If they have x-axis as a common tangent, then:

A) \[\frac{1}{\sqrt{a}}=\frac{1}{\sqrt{b}}+\frac{1}{\sqrt{c}}\]       

B) a, b, c are in A.P

C) \[\sqrt{a},\sqrt{b},\sqrt{c}\] are in A.P D.

D) \[\frac{1}{\sqrt{b}}=\frac{1}{\sqrt{a}}=+\frac{1}{\sqrt{c}}\]

Correct Answer: A

Solution :

\[a<b<c\]are radii of 3 circles, (say \[{{\text{c}}_{\text{1}}}\text{,}\]\[{{\text{c}}_{\text{2}}}\text{,}\]\[{{\text{c}}_{\text{3}}}\text{,}\]) touching each other externally

Also \[x-\]axis is their common tangent then Sum of lengths of common tangents of \[{{\text{c}}_{\text{1}}}\,{{\text{c}}_{\text{2}}}\] these and \[{{\text{c}}_{2}}\,{{\text{c}}_{3}}\]= Lengths of the common tangent of \[{{\text{c}}_{2}}\,{{\text{c}}_{3}}\]

\[\therefore \]\[\sqrt{{{(a+b)}^{2}}\,-\,{{(a-b)}^{2}}+\sqrt{{{(a+c)}^{2}}-{{(a-c)}^{2}}}}\]\[=\,\sqrt{{{(b+c)}^{2}}\,-\,(b-c)}\]

\[\Rightarrow \]   \[\sqrt{ab}+\sqrt{ac}\,=\,\sqrt{bc}\]

\[\Rightarrow \]   \[\frac{1}{\sqrt{c}}\,+\,\frac{1}{\sqrt{b}}\,=\,\frac{1}{\sqrt{a}}.\]