Let the circle S: 36 x ^ 2 + 36 y ^ 2 - 108x + 120y + C = 0 be such...

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

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

We know that the centre and radius of a circle ${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0$ is \left(- g, - f\right)$and$\sqrt{{g}^{2} + {f}^{2} - c}$respectively.</p><p>Thus, centre$e ≡ \left(- g, - f\right) ≡ \left(\dfrac{3}{2}, - \dfrac{10}{6}\right)$and radius$= r = \sqrt{\dfrac{9}{4} + \dfrac{100}{36} - \dfrac{C}{36}}$</p><p><img src="https://cdn-question-pool.Quantrex.app/pyq/jee_main/b4120070-e52d-43c5-b6a4-6d8c0e00c7b0-image.png" style="width: 300px; height: 193px;" /></p><p>Now, this circle neither intersects nor touches the co-ordinate axes, hence, the radius of the circle is less than the absolute value of the co-ordinate with smaller value.</p><p>$⇒ r < \dfrac{3}{2}$</p><p>$\sqrt{\dfrac{9}{4} + \dfrac{100}{36} - \dfrac{C}{36}} < \dfrac{3}{2}$</p><p>$⇒ \dfrac{9}{4} + \dfrac{100}{36} - \dfrac{C}{36} < \dfrac{9}{4}$</p><p>$⇒ C > 100 \ldots \left(1\right)$</p><p>Now, point of intersection of$x - 2y = 4$and$2x - y = 5$is$( 2, - 1 ) ,$which lies inside the circle$S .$</p><p>The point \left({x}_{1}, {y}_{1}\right)$ lies inside a circle ${x}^{2} + {y}^{2} + 2gx + 2fy + c = 0,$ if ${x}_{1}^{2} + {y}_{1}^{2} + 2g{x}_{1} + 2f{y}_{1} + c < 0$

$⇒ {\left(2\right)}^{2} + {\left(- 1\right)}^{2} - 3\left(2\right) + \dfrac{10}{3}\left(- 1\right) + \dfrac{C}{36} < 0$

$⇒ 4 + 1 - 6 - \dfrac{10}{3} + \dfrac{C}{36} < 0$

$⇒ C < 156 \ldots \left(2\right)$

From \left(1\right)$and \left(2\right),$ we have $100 < C < 156$