The value of $k$ , such that the equation $2 x^{2} + 2 y^{2} - 6 x + 8 y + k = 0$ represents a point circle, is

0

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25

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\frac{25}{2}

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- \frac{25}{2}

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Solution

The correct option is B $\frac{25}{2}$
Solution :
$2 x^{2} + 2 y^{2} - 6 x + 8 y + k = 0$
Formula for circle
circle : $(x - a)^{2} + (y - b)^{2} = r^{2}$
writing in above formal,
$x^{2} + y^{2} - 3 x + 4 y + \frac{k}{2} = 0$ (divide by 2)
$x^{2} - 3 x + \frac{9}{4} + y^{2} + 4 y + 4 - \frac{- 9}{4} - 4 + \frac{k}{2} = 0$
$\Rightarrow (x - \frac{3}{2})^{2} + (y + 2)^{2} = \frac{25}{4} - \frac{k}{2}$
For a point circle $\Rightarrow r^{2} = 0$
$∴ \frac{25}{4} - \frac{k}{2} = 0 \Rightarrow k = \frac{25}{2}$

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