If the circle $x^{2} + y^{2} - 6 x - 8 y + (25 - a^{2}) = 0$ touches the axis of $x$ , then $a$ equals to

\pm 4

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\pm 3

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0

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\pm 2

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Solution

The correct option is A $\pm 4$
Equation of circle-
${(x - h)}^{2} + {(y - k)}^{2} = r^{2} ⟶ (i)$

Given equation of circle-

$x^{2} + y^{2} - 6 x - 8 y + (25 - a^{2}) = 0$

$\Rightarrow x^{2} - 6 x + 9 + y^{2} - 8 y + 16 = a^{2}$

$\Rightarrow {(x - 3)}^{2} + {(y - 4)}^{2} = a^{2} ⟶ (i i)$

On comparing ${e q}^{n} (i) & (i i)$ , we get

$h = 3$

$k = 4$

$r = \pm a$

when circle touches $x$ -axis then $r = k = 4$

$∴ a = \pm 4$

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