$$C:4{x^2} + 4{y^2} - 12x + 8y + k = 0$$

$$\because$$ $$\left( {1, - {1 \over 3}} \right)$$ lies on or inside the C

then $$4 + {4 \over 9} - 12 - {8 \over 3} + k \le 0$$

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

Now, circle lies in 4^th quadrant centre $$ \equiv \left( {{3 \over 2}, - 1} \right)$$

$$\therefore$$ $$r < 1 \Rightarrow \sqrt {{9 \over 4} + 1 - {k \over 4}} < 1$$

$$ \Rightarrow {{13} \over 4} - {k \over 4} < 1$$

$$ \Rightarrow {k \over 4} > {9 \over 4}$$

$$ \Rightarrow k > 9$$

$$\therefore$$ $$k \in \left( {9,{{92} \over 9}} \right)$$