question_answer

A chord AB of a circle \[{{C}_{1}}\]of radius \[(\sqrt{3}+1)\,cm\]touches a circle \[{{C}_{2}}\] which is concentric to \[{{C}_{1}}.\] If the radius of \[{{C}_{2}}\] is \[(\sqrt{3}-1)\,cm,\] then the length of AB is

A) \[8\sqrt{3}\,cm\]            

B) \[4\sqrt[4]{3}\,cm\]

C) \[4\sqrt{3}\,cm\]            

D) \[2\sqrt[4]{3}\,cm\]

Correct Answer: B

Solution :

Let the chord AB of circle \[{{C}_{1}}\] touches the circle \[{{C}_{2}}\] at point M.

Then, \[OA=\sqrt{3}+1\]

and \[OM=\sqrt{3}-1\]

Now, in right angled \[\Delta OAM,\]

\[A{{M}^{2}}=O{{A}^{2}}-O{{M}^{2}}\]

\[=\,\,{{(\sqrt{3}+1)}^{2}}-{{(\sqrt{3}-1)}^{2}}\]

\[\Rightarrow \]   \[A{{M}^{2}}=4\sqrt{3}\]

\[\Rightarrow \]   \[AM=2\sqrt[4]{3}\]

\[\therefore \]      \[AB=2AM=4\sqrt[4]{3}\]