question_answer

The circumcircle of \[\Delta ABC\] with \[\angle A={{60}^{o}}\]has its centre at O and radius equal to 2 cm. Circle centered at O., touches the circumcircle and also 08 and OC. The radius of the smaller circle is (in cm)____.

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 \[\cos {{30}^{o}}=\frac{N{{O}_{1}}}{O{{O}_{1}}}\Rightarrow \frac{\sqrt{3}}{2}=\frac{r}{2-r}\Rightarrow 2\sqrt{3}-\sqrt{3}r=2r\]\[=r=\frac{2\sqrt{3}}{2+\sqrt{3}}\,\,\,\Rightarrow \,\,\,r=2\sqrt{3}\,(2-\sqrt{3})=4\sqrt{3}-6\]