Mathematics - Circle Question with Solution | TestHub

$x^2+y^2\le 6$ and $2x-3y=1$ is shown as, For the point to lie in the shaded part, origin and the point lie on opposite side of straight line $L$.
$\therefore$ For any point in shaded part $L>0$ and for any point inside the circle $S<0$. Now, for $\left(2,\frac{3}{4}\right)L:2x-3y-1$ $$L:4-\frac{9}{4}-1=\frac{3}{4}>0$$ 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 $\left(\frac{5}{2},\frac{3}{4}\right)L:5-9-1<0$ [neglect] For $\left(\frac{1}{4},-\frac{1}{4}\right)L:\frac{1}{2}+\frac{3}{4}-1>0\}$ $\therefore \left(\frac{1}{4},-\frac{1}{4}\right)$ 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.