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 unitYou need to login to perform this action.
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