The straight line 2 x-3 y=1 divides the circular region x^2+y^2 ≤ 6...

x^2+y^2 \leq 6$and$2 x-3 y=1$is shown as, For the point to lie in the shaded part, origin and the point lie on opposite side of straight line$L$. <img src="https://cdn-question-pool.Quantrex.app/pyq/jee_advanced/dVwnyTkUUvsDQ18fHYOVS4iDdvsVr-wqh05DruuYZs0.original.fullsize.png"><br>$\therefore$For any point in shaded part$L>0$and for any point inside the circle$S < 0$. Now, for$\left(2, \frac{3}{4}\right) L: 2 x-3 y-1$L: 4-\frac{9}{4}-1=\frac{3}{4}>0$$and$S: x^2+y^2-6$,$$S: 4+\frac{9}{16}-6 < 0$\Rightarrow\left(2 \frac{3}{4}\right)$lies in shaded part. For$\left(\frac{5}{2}, \frac{3}{4}\right) L: 5-9-1 < 0 \quad$[neglect] For \left.\left(\frac{1}{4},-\frac{1}{4}\right) L: \frac{1}{2}+\frac{3}{4}-1>0\right\} \therefore\left(\frac{1}{4},-\frac{1}{4}\right)$ lies in the shaded part. For \left(\frac{1}{8}, \frac{1}{4}\right) L: \frac{1}{4}-\frac{3}{4}-1 < 0$[neglect]$\Rightarrow Only 2 points lie in the shaded part.