The straight line $2 x-3 y=1$ divides the circular region $x^2+y^2 \leq 6$ into two parts. If…

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

Solution

$x^2+y^2 \leq 6$ 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 $L$.
$\therefore$ 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

Asked in: JEE Advanced 2011 (Paper 2)

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