The straight line $2x - 3y = 1$ divides the circular region $x^{2} + y^{2} \le 6$ into two parts. If $S = \{(2, \frac{3}{4}), (\frac{5}{2}, \frac{3}{4}), (\frac{1}{4}, -\frac{1}{4}), (\frac{1}{8}, \frac{1}{4})\}$ then the number of point(s) in S lying inside the smaller part is

$\text{Circle: } x^2 + y^2 \leq 6, \qquad \text{Line: } 2x - 3y = 1$ $S = \left\{ \left(2, \dfrac{3}{4}\right), \left(\dfrac{5}{2}, \dfrac{3}{4}\right), \left(\dfrac{1}{4}, -\dfrac{1}{4}\right), \left(\dfrac{1}{8}, \dfrac{1}{4}\right) \right\}$ $\text{To find: number of points of } S \text{ inside the smaller part}$ $1.\ \text{find which part is smaller}$ $2.\ \text{test each point: inside circle and on that side}$ $3.\ \text{count}$ $d = \dfrac{|2(0)-3(0)-1|}{\sqrt{2^2+3^2}} = \dfrac{1}{\sqrt{13}} \approx 0.277$ $r = \sqrt{6} \approx 2.449, \qquad d \ll r$ $\text{At } O(0,0): \ 2(0)-3(0)-1 = -1 < 0$ $O \text{ lies in } 2x-3y<1 \Rightarrow \text{smaller part is } 2x-3y>1$ $A_{\text{cap}} = R^2\cos^{-1}\!\dfrac{d}{R} - d\sqrt{R^2-d^2} \approx 8.07$ $\dfrac{\pi R^2}{2} = 3\pi \approx 9.42, \qquad 8.07 < 9.42$ $y = \dfrac{2x-1}{3} \Rightarrow x^2 + \left(\dfrac{2x-1}{3}\right)^2 = 6$ $13x^2 - 4x - 53 = 0 \Rightarrow x = \dfrac{2 \pm \sqrt{693}}{13}$ $P_1\left(2,\dfrac{3}{4}\right): \ x^2+y^2 = 4+\dfrac{9}{16} = \dfrac{73}{16} < 6 \ \text{✓}$ $2x-3y = 4-\dfrac{9}{4} = \dfrac{7}{4} > 1 \ \text{✓} \Rightarrow \text{counted}$ $P_2\left(\dfrac{5}{2},\dfrac{3}{4}\right): \ x^2+y^2 = \dfrac{25}{4}+\dfrac{9}{16} = \dfrac{109}{16} \approx 6.81 > 6 \ \text{✗}$ $P_3\left(\dfrac{1}{4},-\dfrac{1}{4}\right): \ x^2+y^2 = \dfrac{1}{16}+\dfrac{1}{16} = \dfrac{1}{8} < 6 \ \text{✓}$ $2x-3y = \dfrac{1}{2}+\dfrac{3}{4} = \dfrac{5}{4} > 1 \ \text{✓} \Rightarrow \text{counted}$ $P_4\left(\dfrac{1}{8},\dfrac{1}{4}\right): \ x^2+y^2 = \dfrac{5}{64} < 6 \ \text{✓}$ $2x-3y = \dfrac{1}{4}-\dfrac{3}{4} = -\dfrac{1}{2} < 1 \ \text{✗ (larger part)}$ $P_1, P_3 \in \text{smaller part}; \quad P_2 \notin \text{circle}; \quad P_4 \in \text{larger part}$ $\boxed{\text{Number of points} = 2}$