The set of values of $$k$$ for which the circle $$C : 4x^2 + 4y^2 - 12x + 8y + k = 0$$ lies inside the fourth quadrant and the point $$(1, -\frac{1}{3})$$ lies on or inside the circle $$C$$ is

The circle is $$C: 4x^2 + 4y^2 - 12x + 8y + k = 0$$. We need to find $$k$$ such that the circle lies inside the fourth quadrant and the point $$(1, -\frac{1}{3})$$ lies on or inside the circle.

Rewrite the circle in standard form

Dividing by 4: $$x^2 + y^2 - 3x + 2y + \frac{k}{4} = 0$$

Completing the square:

$$\left(x - \frac{3}{2}\right)^2 + (y + 1)^2 = \frac{9}{4} + 1 - \frac{k}{4} = \frac{13 - k}{4}$$

Centre: $$\left(\frac{3}{2}, -1\right)$$, Radius: $$r = \frac{\sqrt{13 - k}}{2}$$

Condition for the circle to exist

$$r^2 > 0 \Rightarrow 13 - k > 0 \Rightarrow k < 13$$

Circle lies in the fourth quadrant

The centre is $$\left(\frac{3}{2}, -1\right)$$ which is in the fourth quadrant ($$x > 0, y < 0$$). For the entire circle to lie in the fourth quadrant:

The circle must not cross the x-axis ($$y = 0$$): distance from centre to x-axis $$>$$ radius

$$1 > \frac{\sqrt{13-k}}{2} \Rightarrow 4 > 13 - k \Rightarrow k > 9$$

The circle must not cross the y-axis ($$x = 0$$): distance from centre to y-axis $$>$$ radius

$$\frac{3}{2} > \frac{\sqrt{13-k}}{2} \Rightarrow 9 > 13 - k \Rightarrow k > 4$$

This is already satisfied when $$k > 9$$.

Point $$(1, -\frac{1}{3})$$ lies on or inside the circle

Substituting into the circle equation (the point must satisfy $$\leq 0$$):

$$4(1)^2 + 4\left(\frac{1}{9}\right) - 12(1) + 8\left(-\frac{1}{3}\right) + k \leq 0$$

$$4 + \frac{4}{9} - 12 - \frac{8}{3} + k \leq 0$$

$$\frac{36 + 4 - 108 - 24}{9} + k \leq 0$$

$$\frac{-92}{9} + k \leq 0$$

$$k \leq \frac{92}{9}$$

Combine the conditions

$$k > 9$$ and $$k \leq \frac{92}{9}$$

So $$k \in \left(9, \frac{92}{9}\right]$$

The correct answer is Option D: $$\left(9, \frac{92}{9}\right]$$.