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