For the two circles $$x^2 + y^2 = 16$$ and $$x^2 + y^2 - 2y = 0$$, there is/are:

We are given two circles:

First circle: $$ x^2 + y^2 = 16 $$

Second circle: $$ x^2 + y^2 - 2y = 0 $$

To find the number of common tangents, we first rewrite both equations in standard form to identify their centers and radii.

For the first circle: $$ x^2 + y^2 = 16 $$. This is already in standard form, so it can be written as $$ (x - 0)^2 + (y - 0)^2 = 4^2 $$. Therefore, the center is $$ C_1 = (0, 0) $$ and the radius is $$ r_1 = 4 $$.

For the second circle: $$ x^2 + y^2 - 2y = 0 $$. We complete the square for the $$ y $$-terms. Rewrite as $$ x^2 + (y^2 - 2y) = 0 $$. To complete the square, add and subtract 1: $$ x^2 + (y^2 - 2y + 1) = 1 $$, which simplifies to $$ x^2 + (y - 1)^2 = 1 $$. Therefore, the center is $$ C_2 = (0, 1) $$ and the radius is $$ r_2 = 1 $$.

Now, we have:

• Circle 1: Center $$ (0, 0) $$, Radius $$ 4 $$
• Circle 2: Center $$ (0, 1) $$, Radius $$ 1 $$

Next, we calculate the distance $$ d $$ between the centers $$ C_1(0, 0) $$ and $$ C_2(0, 1) $$:

$$ d = \sqrt{(0 - 0)^2 + (0 - 1)^2} = \sqrt{0 + 1} = \sqrt{1} = 1 $$

We compare this distance with the sum and difference of the radii:

• Sum of radii: $$ r_1 + r_2 = 4 + 1 = 5 $$
• Difference of radii: $$ |r_1 - r_2| = |4 - 1| = 3 $$

The relative positions of the circles determine the number of common tangents:

• If $$ d > r_1 + r_2 $$, the circles are separate and have 4 common tangents (2 direct, 2 transverse).
• If $$ d = r_1 + r_2 $$, the circles touch externally and have 3 common tangents.
• If $$ |r_1 - r_2| < d < r_1 + r_2 $$, the circles intersect at two points and have 2 common tangents.
• If $$ d = |r_1 - r_2| $$, the circles touch internally and have 1 common tangent.
• If $$ d < |r_1 - r_2| $$, one circle lies entirely inside the other without touching, and there are no common tangents.

Here, $$ d = 1 $$ and $$ |r_1 - r_2| = 3 $$. Since $$ 1 < 3 $$, we have $$ d < |r_1 - r_2| $$. This means circle 2 lies entirely inside circle 1 without touching it.

To confirm, we check for points of intersection by solving the equations simultaneously:

Equation of circle 1: $$ x^2 + y^2 = 16 $$

Equation of circle 2: $$ x^2 + y^2 - 2y = 0 $$

Subtract circle 2 from circle 1:

$$ (x^2 + y^2) - (x^2 + y^2 - 2y) = 16 - 0 $$

$$ x^2 + y^2 - x^2 - y^2 + 2y = 16 $$

$$ 2y = 16 $$

$$ y = 8 $$

Substitute $$ y = 8 $$ into circle 1:

$$ x^2 + (8)^2 = 16 $$

$$ x^2 + 64 = 16 $$

$$ x^2 = 16 - 64 = -48 $$

This gives $$ x^2 = -48 $$, which has no real solutions. Therefore, the circles do not intersect and do not touch.

Since circle 2 is entirely inside circle 1 without touching, there are no common tangents. Any tangent to circle 2 (the inner circle) will intersect circle 1 (the outer circle) at two points, as verified by testing tangents at various points on circle 2. For example:

• At point $$ (0, 0) $$ on circle 2, the tangent is $$ y = 0 $$ (x-axis). Substituting into circle 1: $$ x^2 + 0^2 = 16 $$ gives $$ x = \pm 4 $$, so it intersects circle 1 at two points.
• At point $$ (0, 2) $$ on circle 2, the tangent is $$ y = 2 $$. Substituting into circle 1: $$ x^2 + 2^2 = 16 $$ gives $$ x^2 = 12 $$, so $$ x = \pm 2\sqrt{3} $$, again two intersection points.
• At point $$ (1, 1) $$ on circle 2, the tangent is $$ x = 1 $$. Substituting into circle 1: $$ 1^2 + y^2 = 16 $$ gives $$ y^2 = 15 $$, so $$ y = \pm \sqrt{15} $$, two intersection points.

Thus, no line is tangent to both circles.

Hence, the correct answer is Option D.