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

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Circle

The straight line

2 x - 3 y = 1

divides the circular region

x^{2} + y^{2} \leq 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

Answer: 2

The line

L : 2 x - 3 y - 1 = 0

divides the circle

x^{2} + y^{2} = 6

into two parts.

The center of the circle is at the origin. The distance from the origin to the line is:

d = \frac{∣2 ( 0 ) - 3 ( 0 ) - 1∣}{4 + 9} = \frac{1}{13}

Since

d < 6

(the radius), the line intersects the circle.

The smaller part is on the side of the line where

2 x - 3 y - 1 > 0

(away from the origin, since

2 (0) - 3 (0) - 1 = - 1 < 0

).

Checking each point:

$(2, 3/4)$ : $2 (2) - 3 (3/4) - 1 = 4 - 9/4 - 1 = 7/4 > 0$ and $4 + 9/16 = 73/16 < 6$ . Inside smaller part.
$(5/2, 3/4)$ : $2 (5/2) - 3 (3/4) - 1 = 5 - 9/4 - 1 = 7/4 > 0$ and $25/4 + 9/16 = 109/16 > 6$ . Outside circle.
$(1/4, - 1/4)$ : $2 (1/4) - 3 (- 1/4) - 1 = 1/2 + 3/4 - 1 = 1/4 > 0$ and $1/16 + 1/16 = 1/8 < 6$ . Inside smaller part.
$(1/8, 1/4)$ : $2 (1/8) - 3 (1/4) - 1 = 1/4 - 3/4 - 1 = - 3/2 < 0$ . Not in smaller part.

Therefore, 2 points lie inside the smaller part.

JEE Advanced · 2011 · 10 Apr 2011 · Paper 2 · Integer answer