If the circle x² + y² + 2ax + 8y + 16 = 0 touches x-axis, then the value of a is
(a) ± 16
(b) ± 4
(c) ± 8
(d) ± 1

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Solution

(b) ± 4

The equation of the circle is x² + y² + 2ax + 8y + 16 = 0.
Its centre is $(- a, - 4)$ and its radius is a units.

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

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

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