The number of common tangents to the circles $$x^2 + y^2 - 4x - 6y - 12 = 0$$ and $$x^2 + y^2 + 6x + 18y + 26 = 0$$, is

We are given two circles:

Circle 1: $$x^2 + y^2 - 4x - 6y - 12 = 0$$

Circle 2: $$x^2 + y^2 + 6x + 18y + 26 = 0$$

First, let us rewrite each circle in standard form $$(x - h)^2 + (y - k)^2 = r^2$$.

For Circle 1, complete the square:

$$(x^2 - 4x + 4) + (y^2 - 6y + 9) = 12 + 4 + 9 = 25$$

$$(x - 2)^2 + (y - 3)^2 = 25$$

So centre $$C_1 = (2, 3)$$ and radius $$r_1 = 5$$.

For Circle 2, complete the square:

$$(x^2 + 6x + 9) + (y^2 + 18y + 81) = -26 + 9 + 81 = 64$$

$$(x + 3)^2 + (y + 9)^2 = 64$$

So centre $$C_2 = (-3, -9)$$ and radius $$r_2 = 8$$.

Now find the distance between centres:

$$d = \sqrt{(2 - (-3))^2 + (3 - (-9))^2} = \sqrt{5^2 + 12^2} = \sqrt{25 + 144} = \sqrt{169} = 13$$

Check the relationship between $$d$$, $$r_1$$, and $$r_2$$:

$$r_1 + r_2 = 5 + 8 = 13 = d$$

When the distance between centres equals the sum of the radii ($$d = r_1 + r_2$$), the two circles touch each other externally.

For two circles that touch externally, the number of common tangents is 3 (two external tangents and one common tangent at the point of contact).

Therefore, the number of common tangents is 3, which is Option D.