The straight line 2x - 3y = 1 divides the cir | vedclass

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(B) Let the circle be $C: x^2 + y^2 - 6 = 0$ and the line be $L: 2x - 3y - 1 = 0$.First,check which points lie inside the circle $x^2 + y^2 \leq 6$:$1

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$2$

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(B) Let the circle be $C: x^2 + y^2 - 6 = 0$ and the line be $L: 2x - 3y - 1 = 0$.First,check which points lie inside the circle $x^2 + y^2 \leq 6$:$1$. For $(2, 3/4)$: $2^2 + (3/4)^2 = 4 + 9/16 = 73/16 = 4.5625 $2$. For $(5/2, 3/4)$: $(5/2)^2 + (3/4)^2 = 25/4 + 9/16 = 109/16 = 6.8125 > 6$ (Outside).$3$. For $(1/4, -1/4)$: $(1/4)^2 + (-1/4)^2 = 1/16 + 1/16 = 2/16 = 0.125 $4$. For $(1/8, 1/4)$: $(1/8)^2 + (1/4)^2 = 1/64 + 1/16 = 5/64 = 0.078 Now,check which side of the line $L(x, y) = 2x - 3y - 1$ these points lie on. The center of the circle $(0, 0)$ gives $L(0, 0) = -1 $1$. For $(2, 3/4)$: $L = 2(2) - 3(3/4) - 1 = 4 - 9/4 - 1 = 3 - 2.25 = 0.75 > 0$.$2$. For $(1/4, -1/4)$: $L = 2(1/4) - 3(-1/4) - 1 = 1/2 + 3/4 - 1 = 1.25 - 1 = 0.25 > 0$.$3$. For $(1/8, 1/4)$: $L = 2(1/8) - 3(1/4) - 1 = 1/4 - 3/4 - 1 = -1.5 Since $L(0, 0) = -1$,the region containing the origin is $L  0$. The points $(2, 3/4)$ and $(1/4, -1/4)$ satisfy $L > 0$ and are inside the circle. Thus,there are $2$ such points.

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