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

C: $4x^2+4y^2-12x+8y+k=0$
$\because (1,-\frac{1}{3})$ lies on or inside the $C$
then $4+\frac{4}{9}-12-\frac{8}{3}+k\le 0$

$$\Rightarrow k\le \frac{92}{9}$$

Now, circle lies in $4^{\text{th }}$ quadrant centre $\equiv (\frac{3}{2},-1)$
$\therefore r<1\Rightarrow \sqrt{\frac{9}{4}+1-\frac{k}{4}}<1$

$\begin{array}{rl} & \Rightarrow \frac{13}{4}-\frac{k}{4}<1\\ & \Rightarrow \frac{k}{4}>\frac{9}{4}\\ & \Rightarrow k>9\\ & \therefore k\in (9,\frac{92}{9}]\end{array}$