Animated Solution for Mathematics - Circles: The straight line $2x-3y=1$ divides the circular region $x^2+y^2\le 6$ into two parts. If $S=\left\{(2,3/4),(5/2,3/4),(1/4,-1/4),(1/8,1/4)\right\}$ then the number of points (s) in $S$ lying inside the smaller part is

Imagine you are standing on a vast, flat plane. In front of you lies a circular region defined by the inequality $x^2+y^2\le 6$.

This is our territory. The center of this circle is anchored at the origin, $(0,0)$, and its radius is $\sqrt{6}$, which is approximately $2.45$. This circle is the boundary of our world; any point we consider must first prove it belongs within this $2.45$-unit radius.

Now, a straight line, $2x-3y=1$, cuts through this circular region like a blade. It doesn't pass through the center, which means it doesn't bisect the circle into two equal halves. Instead, it creates two unequal pieces: a larger one and a smaller one.

Our mission is to identify which of our candidate points from the set $S=\{(2,3/4),(5/2,3/4),(1/4,-1/4),(1/8,1/4)\}$ reside within that elusive, smaller piece.

To conquer this, we need a way to distinguish the 'smaller' side from the 'larger' side without getting lost in complex sketches. We define our line function as $L(x,y)=2x-3y-1$.

This function is our compass. If we plug the origin $(0,0)$ into this function, we get $L(0,0)=2(0)-3(0)-1=-1$. Since $-1<0$, we know that the origin sits on the side of the line where the expression is negative.

But wait—is this the smaller side or the larger side? The perpendicular distance from the center $(0,0)$ to the line is:

Since $0.28$ is much smaller than the radius $2.45$, the line is very close to the center. The origin is clearly in the larger part of the circle.

Therefore, the smaller part must be the region where $L(x,y)>0$. This is our golden rule: a point is in the smaller part if and only if $2x-3y-1>0$.

Now, we test our candidates. A point is a winner only if it passes two tests: it must be inside the circle ($x^2+y^2\le 6$) AND it must be in the smaller part ($2x-3y-1>0$).

Circle test: $2^2+(3/4{)}^2=4+0.5625=4.5625$. Since $4.5625\le 6$, it is inside the circle.

Line test: $2(2)-3(3/4)-1=4-2.25-1=0.75$. Since $0.75>0$, it is in the smaller part. $P_1$ is a winner!

Circle test: $(5/2{)}^2+(3/4{)}^2=6.25+0.5625=6.8125$. Since $6.8125>6$, it is outside the circle. $P_2$ is out.

Circle test: $(1/4{)}^2+(-1/4{)}^2=1/16+1/16=0.125$. Since $0.125\le 6$, it is inside the circle.

Line test: $2(1/4)-3(-1/4)-1=0.5+0.75-1=0.25$. Since $0.25>0$, it is in the smaller part. $P_3$ is a winner!

Circle test: $(1/8{)}^2+(1/4{)}^2=1/64+4/64=5/64\approx 0.078$. Since $0.078\le 6$, it is inside the circle.

Line test: $2(1/8)-3(1/4)-1=0.25-0.75-1=-1.5$. Since $-1.5<0$, it is in the larger part. $P_4$ is out.

After our rigorous testing, we find that only $P_1$ and $P_3$ satisfy both conditions. The beauty of this problem lies in the simplicity of the algebraic test.

By establishing the sign of the region using the origin, we turned a complex geometric visualization into a simple, logical verification. There are exactly $2$ points in the smaller part.

Keep this systematic approach in your toolkit—it is the key to solving even the most daunting coordinate geometry problems.