The power of (2,1) with respect to the circle $2 x^{2} + 2 y^{2} - 8 x - 6 y + k$ = 0 is positive if

0<k<12

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-12<k<12

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k>12

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k<12

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Solution

The correct option is C
k>12

$2 x^{2} + 2 y^{2} - 8 x - 6 y + k$ = 0

$\Rightarrow x^{2} + y^{2} - 4 x - 3 y + (\frac{k}{2})$ = 0

$4 + 1 - 8 - 3 + \frac{k}{2} > 0$

$\frac{k}{2} - 6 > 0 \to k > 12$

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