Radius of the circle $2 x^{2} + 2 y^{2} + 8 x + 4 y - 3 = 0$ is

\sqrt{17}

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\sqrt{23}

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\sqrt{\frac{13}{2}}

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\sqrt{\frac{11}{2}}

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Solution

The correct option is B $\sqrt{23}$
Formula,

$r = \sqrt{g^{2} + f^{2} - c}$

From given, we have,

$2 g = 8 \Rightarrow g = 4$

$2 f = 4 \Rightarrow f = 2$

$c = - 3$

$r = \sqrt{4^{2} + 2^{2} - (- 3)} = \sqrt{23}$

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