question_answer

Circle \[{{C}_{2}}\] passes through the centre of circle \[{{C}_{1}}\]and is tangential to it. If the area of \[{{C}_{1}}\]is \[4\text{ }c{{m}^{2}},\] then the area of \[{{C}_{2}}\] is _____.

A)  \[8\,c{{m}^{2}}\]                           

B)  \[8\,\sqrt{\pi \,}c{{m}^{2}}\]         

C)  \[16\,c{{m}^{2}}\]       

D)         \[16\sqrt{\pi }\,c{{m}^{2}}\]

Correct Answer: C

Solution :

Given figure is                            Area of \[{{C}_{1}}=4\,c{{m}^{2}}\] \[\Rightarrow \]            \[4=\pi {{r}_{1}}^{2}\] \[\Rightarrow \]            \[\sqrt{\frac{4}{\pi }}={{r}_{1}}\] Now,   \[{{r}_{2}}=2{{r}_{1}}\] \[\Rightarrow \]            \[{{r}_{2}}=\frac{2\times 2}{\sqrt{\pi }}=\frac{4}{\sqrt{\pi }}\] Area of \[{{C}_{2}}=\pi \times {{r}_{2}}^{2}=\pi \times \frac{4\times 4}{\pi }=16c{{m}^{2}}\]