question_answer

Two circles touch externally. The sum of their areas is \[130\pi \,c{{m}^{2}}\] and the distance between their centres is 14 cm. The radius of the smaller circle is                                                                         [SSC (CGL) Pre 2015]

A) 2 cm                            

B) 3 cm

C) 4 cm                            

D) 5 cm

Correct Answer: B

Solution :

Here,

Now, \[\pi \,{{(14-R)}^{2}}+\pi \,{{R}^{2}}=130\pi \]

\[196+{{R}^{2}}-28R+{{R}^{2}}=130\]

\[\Rightarrow \]   \[2{{R}^{2}}-28R+66=0\]

\[\Rightarrow \]   \[{{R}^{2}}-14R+33=0\]

\[\Rightarrow \]   \[{{R}^{2}}-11R-3R+33=0\]

\[\Rightarrow \]   \[R\,(R-11)-3\,(R-11)=0\]\[\Rightarrow \]\[R=11\]

\[\therefore \] Radius of smaller circle \[=14-11=3\,cm\]