The given equation is:
$$|z - 4| = |z + 4|$$
Geometrically, this represents the set of all points $$z$$ in the complex plane that are equidistant from the fixed points $$(4, 0)$$ and $$(-4, 0)$$. The locus of such points is the perpendicular bisector of the line segment joining $$(4, 0)$$ and $$(-4, 0)$$, which is the y-axis ($$x = 0$$).
Alternatively, we can substitute $$z = x + iy$$ to solve algebraically:
$$|x + iy - 4| = |x + iy + 4|$$ $$\sqrt{(x - 4)^2 + y^2} = \sqrt{(x + 4)^2 + y^2}$$
Square both sides to eliminate the square roots:
$$(x - 4)^2 + y^2 = (x + 4)^2 + y^2$$ $$x^2 - 8x + 16 = x^2 + 8x + 16$$ $$-8x = 8x \implies 16x = 0 \implies x = 0$$
Thus, the locus line is the line $$x = 0$$.
We need to find the points on this line that lie strictly inside the circle $$|z| = 6$$.
The equation of the circle centered at the origin with radius $$R = 6$$ is:
$$x^2 + y^2 < 6^2 \implies x^2 + y^2 < 36$$
Since the points must lie on our locus line ($$x = 0$$), substitute $$x = 0$$ into the inequality:
$$0^2 + y^2 < 36$$ $$y^2 < 36 \implies -6 < y < 6$$
A lattice point must have integer coordinates. Since $$x = 0$$, we just need to count the integer values of $$y$$ that satisfy the inequality $$-6 < y < 6$$:
The valid integer values for $$y$$ are:
$$\{-5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5\}$$
Counting these values, we find there are exactly 11 valid integer values for $$y$$.
Therefore, there are $$11$$ lattice points on the line inside the circle, matching Option A.
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