If 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:

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$-12

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k > 12

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k < 12

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Solution

The correct option is D $k > 12$
$\begin{matrix} 2 x^{2} + 2 y^{2} - 8 x - 6 y + k = 0 x^{2} + y^{2} - 4 x - 3 y + \frac{k}{2} = 0 {(2)}^{2} + {(1)}^{2} - 4 (2) - 3 (1) + \frac{k}{2} > 0 \frac{k}{2} - 6 > 0 k > 12 H e n c e, t h e o p t i o n C i s t h e c o r r e c t a n s w e r . \end{matrix}$

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Similar questions

If 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

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

(2, 1)

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

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

(2, 1)

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

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