If the circle $x^{2} + y^{2} + 2 a x + 8 y + 16 = 0$
touches x-axis, then the value of a is

$\pm 16$

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$\pm 4$

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$\pm 8$

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$\pm 1$

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Solution

The correct option is B
$\pm 4$

The equation of the circle is

$x^{2} + y^{2} + 2 a x + 8 y + 16 = 0$

Its centre is (-a, -4) and its radius is a units.

Since the circle touches the x-axis, we have :

$= \sqrt{(- a + a)^{2} + (4 - 0)^{2}} = a$

$\Rightarrow a = \pm 4$

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