Let the circle 

$S:36x^2+36y^2-108x+120y+C=0$Be such that it neither intersects nor touches the co-ordinate axes. If the point of intersection of the lines, x – 2y = 4 and 2x – y = 5 lies inside the circle S, then: 

 

$$\begin{gathered} \mathrm{S}: 36{\mathrm{x}}^2+36{\mathrm{y}}^2-108\mathrm{x}+120\mathrm{y}+C=0 \\ Given lines \\ \mathrm{x}-2\mathrm{y}-4=0 \\ 2\mathrm{x}-\mathrm{y}-5=0 \\ \mathrm{whose} \mathrm{point} \mathrm{of} \mathrm{inter} \mathrm{section} \mathrm{is} (2, -1) which \mathrm{is} \mathrm{inside} \mathrm{the} \mathrm{circle} \mathrm{then} {\mathrm{S}}_{11}<0 \\ \therefore 36\left(4\right)+36\left(1\right)-108\left(2\right)+120(-1)+C<0 \\ \Rightarrow C<156\cdots \left(i\right) \\ and {\left(\frac{108}{72}\right)}^2+{\left(\frac{-120}{72}\right)}^2-\frac{C}{36}<{\left(\frac{3}{2}\right)}^2 (Neither touches any axis) \\ \Rightarrow \frac{9}{4}+\frac{25}{9}-\frac{C}{36}<\frac{9}{4} \\ \therefore 100<C\cdots \left(ii\right) \\ From \left(i\right) and \left(ii\right) 100<C<156 \end{gathered}$$