Question: 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
Options:
A) $ 8\sqrt{3},cm $
B) $ 4\sqrt[4]{3},cm $
C) $ 4\sqrt{3},cm $
D) $ 2\sqrt[4]{3},cm $
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Answer:
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, $
$ AM^{2}=OA^{2}-OM^{2} $
$ ={{(\sqrt{3}+1)}^{2}}-{{(\sqrt{3}-1)}^{2}} $
$ \Rightarrow $ $ AM^{2}=4\sqrt{3} $
$ \Rightarrow $ $ AM=2\sqrt[4]{3} $
$ \therefore $ $ AB=2AM=4\sqrt[4]{3} $