The centres of those circles which touch the circle, $$x^2 + y^2 - 8x - 8y - 4 = 0$$, externally and also touch the x-axis, lie on

We consider the fixed circle whose equation is $$x^{2}+y^{2}-8x-8y-4=0.$$

First we rewrite this equation in standard form. We group the $$x$$ and $$y$$ terms and complete the squares:

$$x^{2}-8x+y^{2}-8y=4$$

$$x^{2}-8x+16+y^{2}-8y+16=4+16+16$$

$$(x-4)^{2}+(y-4)^{2}=36.$$

Hence the centre of the given circle is $$(4,\,4)$$ and its radius is $$R=6.$$

Now let $$C(h,k)$$ be the centre of a variable circle that satisfies the two required conditions.

Condition 1 - Touching the x-axis. A circle with centre $$(h,k)$$ touches the x-axis if the perpendicular distance of its centre from the x-axis equals its radius. The distance of $$(h,k)$$ from the x-axis $$y=0$$ is $$|k|.$$ Because we are interested in the circle lying above the x-axis, we have $$k>0$$, and therefore the radius of the variable circle is $$r=k.$$

Condition 2 - Touching the fixed circle externally. The distance between the centres equals the sum of the radii. Using the distance formula,

$$\sqrt{(h-4)^{2}+(k-4)^{2}} = r+R \;=\; k+6.$$

We square both sides to eliminate the square root:

$$(h-4)^{2}+(k-4)^{2} = (k+6)^{2}.$$

We now expand each bracket.

Left-hand side:

$$(h-4)^{2} = h^{2}-8h+16,$$

$$(k-4)^{2} = k^{2}-8k+16,$$

so $$(h-4)^{2}+(k-4)^{2}=h^{2}-8h+k^{2}-8k+32.$$

Right-hand side:

$$(k+6)^{2}=k^{2}+12k+36.$$

Equating the two expansions, we have

$$h^{2}-8h+k^{2}-8k+32 = k^{2}+12k+36.$$

The term $$k^{2}$$ appears on both sides, so it cancels:

$$h^{2}-8h-8k+32 = 12k+36.$$

We collect like terms on the left:

$$h^{2}-8h-8k-12k+32-36 = 0,$$

$$h^{2}-8h-20k-4 = 0.$$

This is the required relation between $$h$$ and $$k$$. Because $$h$$ represents the x-coordinate and $$k$$ the y-coordinate of the centre, we now recognise it as the equation of the locus in the usual $$x,y$$ notation:

$$x^{2}-8x-20y-4 = 0.$$

To recognise the conic, we gather the $$x$$ terms and move the remaining terms to the right:

$$x^{2}-8x = 20y+4.$$

We complete the square in $$x$$:

$$x^{2}-8x+16 = 20y+4+16,$$

$$(x-4)^{2} = 20(y+1).$$

This is the standard form $$(x-h_{0})^{2}=4a(y-k_{0})$$ of a parabola whose axis is parallel to the positive $$y$$-direction. Specifically, the parabola has vertex $$(4,\,-1)$$ and focal length $$a=5.$$ Because the equation contains only one squared variable ($$x^{2}$$) and not both $$x^{2}$$ and $$y^{2}$$, the conic cannot be a circle, ellipse (other than a parabola), or hyperbola. It is unequivocally a parabola.

Hence, the correct answer is Option B.