The power of (2,1) with respect to the circle 2x2+2y2−8x−6y+k = 0

Question

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

Options:

A . 0

B . -12

C . k>12

D . k

Answer: Option C
:
C

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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