The radius of the circle $x^{2} + y^{2} + 4 x + 6 y + 13 = 0$ is

\sqrt{26}

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\sqrt{13}

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\sqrt{23}

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0

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Solution

The correct option is D $0$
Given equation is
$x^{2} + y^{2} + 4 x + 6 y + 13 = 0$
or $(x^{2} + 4 x + 4) + (y^{2} + 6 y + 9) + 13 = 4 + 9$
or $(x + 2)^{2} (y + 3)^{2} = 0$
$∴$ Radius of circle $= 0$ .

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