A) \[12-4\sqrt{3}\]
B) \[\frac{6-\sqrt{3}}{2}\]
C) \[\frac{3-\sqrt{3}}{2}\]
D) \[4\sqrt{3}-6\]
Correct Answer: D
Solution :
In \[\Delta ABO,\,\,\angle BAO={{30}^{o}}\] And \[\angle ABO={{30}^{o}}\] [since \[AO=OB,\]radii of circle] \[\angle ABO+\angle BAO+\angle AOB={{180}^{o}}\] \[\therefore \] \[\angle AOB={{120}^{o}}\] \[\Rightarrow \] \[\angle BOM={{60}^{o}}\] \[(\because \,\,\angle AOB+\angle BOM={{180}^{o}})\] Let \[{{O}_{1}}M=r,\] Then \[O{{O}_{1}}=2-r\] and \[N{{O}_{1}}=r\] \[N{{O}_{1}}\] is perpendicular to OB, since OB is a tangent to the smaller circle, thus \[\cos {{30}^{o}}=\frac{N{{O}_{1}}}{O{{O}_{1}}}\Rightarrow \frac{\sqrt{3}}{2}=\frac{r}{2-r}\Rightarrow 2\sqrt{3}-\sqrt{3}r=2r\]\[=r=\frac{2\sqrt{3}}{2+\sqrt{3}}\,\,\,\Rightarrow \,\,\,r=2\sqrt{3}\,(2-\sqrt{3})=4\sqrt{3}-6\]You need to login to perform this action.
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