If x2+y2=k2 (where x, y are variables) and k is constant) then x∉y can be takes as x=k cos θ, y= k sin θ and (x−1)2+(y−2)2=4 then maximum value of x2+y2 is equal to

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If $x^{2} + y^{2} = k^{2}$ (where x, y are variables) and k is constant) then $x \notin y$ can be takes as $x = k \cos θ, y = k \sin θ a n d {(x - 1)}^{2} + {(y - 2)}^{2} = 4$ then maximum value of $x^{2} + y^{2}$ is equal to

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$9 + 4 \sqrt{5}$

$1 - 4 \sqrt{5}$

$5 + 2 \sqrt{3}$

$3 + 4 \sqrt{5}$

answer is C.

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Detailed Solution

${(x - 1)}^{2} + {(y - 2)}^{2} = 2^{2} ∴ x = 1 + 2 \cos θ, y = 2 + 2 \sin θ \Rightarrow x^{2} + y^{2} = 9 + 4 \cos θ + 8 \sin θ$
Maximum value $= 9 + \sqrt{16 + 64} = 9 + 4 \sqrt{5}$