1. Write the circle in standard form

Given $4x^2+4y^2-12x+8y+k=0$ Divide by $4$: $x^2+y^2-3x+2y+\frac{k}{4}=0$

Complete the squares: $x^2-3x=\left(x-\frac32\right)^2-\frac94$ $y^2+2y=(y+1)^2-1$

So, $\left(x-\frac32\right)^2+(y+1)^2=\frac{13-k}{4}$

Hence the circle has

• centre $\left(\frac32,-1\right)$
• radius $r=\frac{\sqrt{13-k}}{2}$

For a real circle, we need $13-k>0 \implies k<13$

2. Condition: circle lies inside the fourth quadrant

The fourth quadrant means points satisfy $x>0,\quad y<0$

For the whole circle to lie inside the fourth quadrant, the centre-to-axis distances must be at least the radius, and since it should not cross the axes, we need strict inequalities:

• distance from centre to $y$-axis $=\frac32$ $r<\frac32$
• distance from centre to $x$-axis $=1$ $r<1$

The stronger condition is $r<1$

Thus, $\frac{\sqrt{13-k}}{2}<1$ $\sqrt{13-k}<2$ $13-k<4$ $k>9$

So from this condition, $k>9$

3. Condition: point $\left(1,-\frac13\right)$ lies on or inside the circle

Distance squared of the point from the centre $\left(\frac32,-1\right)$: $d^2=\left(1-\frac32\right)^2+\left(-\frac13+1\right)^2$ $=\left(-\frac12\right)^2+\left(\frac23\right)^2$ $=\frac14+\frac49$ $=\frac{9+16}{36}=\frac{25}{36}$

For the point to lie on or inside the circle: $d^2\le r^2$ $\frac{25}{36}\le \frac{13-k}{4}$ Multiply by $36$: $25\le 9(13-k)$ $25\le 117-9k$ $9k\le 92$ $k\le \frac{92}{9}$

4. Combine both conditions

From quadrant condition: $k>9$ From point condition: $k\le \frac{92}{9}$

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

5. Match with options

This is Option D.

So the required set is $\boxed{\left(9,\frac{92}{9}\right]}$