A point inside the circle x^2 + y^2 + 3x - 3y | vedclass

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(B) Let the equation of the circle be $S(x, y) = x^2 + y^2 + 3x - 3y + 2 = 0$.$A$ point $(x_1, y_1)$ lies inside the circle if $S(x_1, y_1) < 0$.Check

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$(-2, 1)$

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(B) Let the equation of the circle be $S(x, y) = x^2 + y^2 + 3x - 3y + 2 = 0$.$A$ point $(x_1, y_1)$ lies inside the circle if $S(x_1, y_1) Checking option $(A) (-1, 3)$: $S(-1, 3) = (-1)^2 + (3)^2 + 3(-1) - 3(3) + 2 = 1 + 9 - 3 - 9 + 2 = 0$. The point lies on the circle.Checking option $(B) (-2, 1)$: $S(-2, 1) = (-2)^2 + (1)^2 + 3(-2) - 3(1) + 2 = 4 + 1 - 6 - 3 + 2 = -2$. Since $-2 Checking option $(C) (2, 1)$: $S(2, 1) = (2)^2 + (1)^2 + 3(2) - 3(1) + 2 = 4 + 1 + 6 - 3 + 2 = 10$. Since $10 > 0$,the point lies outside the circle.Checking option $(D) (-3, 2)$: $S(-3, 2) = (-3)^2 + (2)^2 + 3(-3) - 3(2) + 2 = 9 + 4 - 9 - 6 + 2 = 0$. The point lies on the circle.Thus,the correct option is $(B)$.

$A$ point inside the circle $x^2 + y^2 + 3x - 3y + 2 = 0$ is

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