Let the circle

Solution:

$S: 36 x^{2}+36 y^{2}-108 x+120 y+C=0$

$\Rightarrow x^{2}+y^{2}-3 x+\frac{10}{3} y+\frac{C}{36}=0$

Centre $\equiv(-g,-f) \equiv\left(\frac{3}{2}, \frac{-10}{6}\right)$

radius $=r=\sqrt{\frac{9}{4}+\frac{100}{36}-\frac{C}{36}}$

Now,

$\Rightarrow \mathrm{r}<\frac{3}{2}$

$\Rightarrow \frac{9}{4}+\frac{100}{36}-\frac{\mathrm{C}}{36}<\frac{9}{4}$

$\Rightarrow \mathrm{C}>100$ .........(1)

Now point of intersection of $x-2 y=4$ and $2 x-y=5$ is $(2,-1)$, which lies inside the circle $S .$

$\therefore \mathrm{S}(2,-1)<0$

$\Rightarrow(2)^{2}+(-1)^{2}-3(2)+\frac{10}{3}(-1)+\frac{C}{36}<0$

$\Rightarrow 4+1-6-\frac{10}{3}+\frac{C}{36}<0$

$\mathrm{C}<156$ ..........(2)

From (1) \& (2)

$100<\mathrm{C}<156$ Ans.