The set of values of k for which the circle C : 4 x 2 + 4 y 2 - 12 x + 8 y + k = 0 lies inside the fourth…

Given equation of circle $C:4x^2+4y^2-12x+8y+k=0$

$\Rightarrow x^2+y^2-3x+2y+\left(\frac{k}{4}\right)=0$

Centre $\left(\frac{3}{2},-1\right);r=\sqrt{\frac{13-k}{2}}\Rightarrow k\le 13 \dots \left(1\right)$

case (i) Point $\left(1,\frac{-1}{3}\right)$ lies on or inside circle $C$

$\Rightarrow S_1\le 0\Rightarrow k\le \frac{92}{9}$    $\cdots \left(2\right)$

case (ii) $C$ lies in $4^{\mathrm{th}}$ quadrant

$r<1$

$\Rightarrow \frac{\sqrt{13-k}}{2}<1$

$\Rightarrow k<9 \cdots \left(3\right)$

Hence from equation $\left(1\right)\cap \left(2\right)\cap \left(3\right)\Rightarrow k\in (9,\frac{92}{9}]$