The equation $k (x^{2} + y^{2}) - x - y + k = 0$ represents a real circle, if

k < \sqrt{2}

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k > \sqrt{2}

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k > 1 / \sqrt{2}

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0 < | K | \leq \frac{1}{\sqrt{2}}

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Solution

The correct option is D $k < \sqrt{2}$
For a circle $x^{2} + y^{2} + 2 g x + 2 f y + c = 0$ to represent a real circle $\sqrt{g^{2} + f^{2} - c} > 0$

Given equation $k (x^{2} + y^{2}) - x - y + k = 0 \Rightarrow (x^{2} + y^{2}) - \frac{1}{k} (x + y) + 1 = 0$

$\sqrt{g^{2} + f^{2} - c} = \sqrt{\frac{1}{k^{2}} + \frac{1}{k^{2}} - 1} = \sqrt{\frac{2}{k^{2}} - 1} > 0$

$\Rightarrow k^{2} < 2 \Rightarrow k < \sqrt{2}$

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