question_answer

If the length of the tangent from any point on the circle \[{{(x-3)}^{2}}+{{(y+2)}^{2}}=5{{r}^{2}}\] to the circle \[{{(x-3)}^{2}}+{{(y+2)}^{2}}={{r}^{2}}\] is 16 unit, then the area between the two circles in sq unit is:

A)  \[32\pi \]                           

B)  \[4\pi \]

C)  \[8\pi \]                                              

D)  \[256\pi \]

Correct Answer: D

Solution :

Let point \[P({{x}_{1}},{{y}_{1}})\] be any point on the circle, therefore it satisfy the circle \[{{({{x}_{1}}-3)}^{2}}+{{({{y}_{1}}+2)}^{2}}=5{{r}^{2}}\]             ?.(i) The length of the tangent drawn from point \[P({{x}_{1}},{{y}_{1}})\] to the circle \[{{(x-3)}^{2}}+{{(y+2)}^{2}}={{r}^{2}}\] is                 \[\sqrt{{{({{x}_{1}}-3)}^{2}}+{{({{y}_{1}}+2)}^{2}}-{{r}^{2}}}\]                                 \[=\sqrt{5{{r}^{2}}-{{r}^{2}}}\]   (From (i))                 \[\Rightarrow \]               \[16=2r\]                 \[\Rightarrow \]               \[r=8\] \[\therefore \] The area between two circles                                   \[=\pi \,5{{r}^{2}}-\pi {{r}^{2}}\]                 \[=4\pi {{r}^{2}}=4\pi \times {{8}^{2}}\]                 \[=256\pi \,\,sq\,\,unit\]