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

The given circle equation is $$ x^2 + y^2 - 6x - 8y + (25 - a^2) = 0 $$. For the circle to touch the x-axis, the distance from its center to the x-axis must equal the radius.

First, rewrite the equation in standard form. The general circle equation is $$ x^2 + y^2 + 2gx + 2fy + c = 0 $$, with center $$(-g, -f)$$ and radius $$ \sqrt{g^2 + f^2 - c} $$. Comparing coefficients:

From $$ -6x = 2gx $$, so $$ 2g = -6 $$, giving $$ g = -3 $$.

From $$ -8y = 2fy $$, so $$ 2f = -8 $$, giving $$ f = -4 $$.

The constant term is $$ c = 25 - a^2 $$.

Thus, the center is $$ (-g, -f) = (3, 4) $$.

Now, the radius $$ r $$ is:

$$ r = \sqrt{g^2 + f^2 - c} = \sqrt{(-3)^2 + (-4)^2 - (25 - a^2)} $$

$$ r = \sqrt{9 + 16 - 25 + a^2} $$

$$ r = \sqrt{25 - 25 + a^2} $$

$$ r = \sqrt{a^2} = |a| $$ (since radius is non-negative).

The distance from the center $$(3, 4)$$ to the x-axis (where $$ y = 0 $$) is the absolute value of the y-coordinate, which is $$ |4| = 4 $$. For the circle to touch the x-axis, this distance must equal the radius:

$$ |a| = 4 $$

Thus, $$ a = 4 $$ or $$ a = -4 $$, so $$ a = \pm 4 $$.

Alternatively, substitute $$ y = 0 $$ into the circle equation to find the condition for tangency with the x-axis:

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

$$ x^2 - 6x + 25 - a^2 = 0 $$

This quadratic in $$ x $$ must have exactly one solution (discriminant zero) for tangency:

Discriminant $$ D = b^2 - 4ac = (-6)^2 - 4(1)(25 - a^2) $$

$$ D = 36 - 4(25 - a^2) $$

$$ D = 36 - 100 + 4a^2 $$

$$ D = 4a^2 - 64 $$

Set $$ D = 0 $$:

$$ 4a^2 - 64 = 0 $$

$$ 4a^2 = 64 $$

$$ a^2 = 16 $$

$$ a = \pm 4 $$

Both methods confirm $$ a = \pm 4 $$. Comparing with the options:

A. 0

B. $$ \pm 4 $$

C. $$ \pm 2 $$

D. $$ \pm 3 $$

Hence, the correct answer is Option B.