If a circle C passing through (4, 0) touches the circle $$x^2 + y^2 + 4x - 6y - 12 = 0$$ externally at a point (1, -1), then the radius of the circle C is :

First, recall the standard form of a circle:

$$x^{2}+y^{2}+2gx+2fy+c=0,$$

where the centre is $$(-g,\,-f)$$ and the radius is $$\sqrt{\,g^{2}+f^{2}-c\,}.$$

For the given circle

$$x^{2}+y^{2}+4x-6y-12=0,$$

we identify $$2g=4 \Rightarrow g=2$$ and $$2f=-6 \Rightarrow f=-3,$$ while $$c=-12.$$ Hence,

centre $$O_{1}=(-g,\,-f)=(-2,\,3),$$

radius

$$r_{1}=\sqrt{g^{2}+f^{2}-c}=\sqrt{2^{2}+(-3)^{2}-(-12)}=\sqrt{4+9+12}=\sqrt{25}=5.$$

The required circle, say $$C,$$ has centre $$O_{2}=(h,k)$$ and radius $$r_{2}.$$ It satisfies three conditions:

1. It passes through the point $$(4,0):$$

$$\boxed{(4-h)^{2}+k^{2}=r_{2}^{2}}\qquad(1)$$

2. It also passes through the point of external tangency $$(1,-1):$$

$$\boxed{(h-1)^{2}+(k+1)^{2}=r_{2}^{2}}\qquad(2)$$

3. Because the circles touch externally at $$(1,-1),$$ the distance between their centres equals the sum of their radii:

$$\sqrt{(h+2)^{2}+(k-3)^{2}} = r_{1}+r_{2}=5+r_{2}.$$

Squaring both sides gives

$$\boxed{(h+2)^{2}+(k-3)^{2}=(5+r_{2})^{2}}\qquad(3)$$

We now proceed algebraically. From (1) and (2) the right-hand sides are equal, so

$$(h-1)^{2}+(k+1)^{2}=(4-h)^{2}+k^{2}.$$

Expanding every term,

$$\bigl(h^{2}-2h+1\bigr)+\bigl(k^{2}+2k+1\bigr)=\bigl(16-8h+h^{2}\bigr)+k^{2}.$$

Cancelling the common $$h^{2}$$ and $$k^{2},$$ we obtain

$$-2h+2k+2=16-8h.$$

Rearranging,

$$6h+2k-14=0 \;\;\Longrightarrow\;\; k=7-3h.\qquad(4)$$

Next, from (2) we write

$$r_{2}^{2}=(h-1)^{2}+(k+1)^{2}.$$

Using (4), $$k+1=(7-3h)+1=8-3h,$$ hence

$$r_{2}^{2}=(h-1)^{2}+(8-3h)^{2}=h^{2}-2h+1+64-48h+9h^{2}=10h^{2}-50h+65.\qquad(5)$$

Now substitute $$k=7-3h$$ into the left side of (3):

$$(h+2)^{2}+(k-3)^{2}=(h+2)^{2}+\bigl((7-3h)-3\bigr)^{2}=(h+2)^{2}+(4-3h)^{2}.$$

Expanding,

$$(h^{2}+4h+4)+(16-24h+9h^{2})=10h^{2}-20h+20.\qquad(6)$$

The right side of (3) is

$$(5+r_{2})^{2}=25+10r_{2}+r_{2}^{2}.$$

Equating (6) with this and recalling (5) for $$r_{2}^{2},$$ we get

$$10h^{2}-20h+20 = 25 + 10r_{2} + (10h^{2}-50h+65).$$

The $$10h^{2}$$ terms cancel, leaving

$$-20h+20 = -50h + 90 + 10r_{2}.$$

Simplifying,

$$30h-70=10r_{2} \;\;\Longrightarrow\;\; r_{2}=3h-7.\qquad(7)$$

Because $$r_{2}$$ is also given by (5), we square (7):

$$(3h-7)^{2}=9h^{2}-42h+49.$$

Set this equal to (5):

$$9h^{2}-42h+49 = 10h^{2}-50h+65.$$

Bringing all terms to one side,

$$h^{2}-8h+16=0 \;\;\Longrightarrow\;\; (h-4)^{2}=0 \;\;\Longrightarrow\;\; h=4.$$

Substituting $$h=4$$ into (7),

$$r_{2}=3(4)-7=12-7=5.$$

Thus the radius of the required circle is

$$r_{2}=5.$$

Hence, the correct answer is Option A.