If $S = x^{2} + y^{2} + x - y - 2 = 0,$ then

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Solution

Let $S (x, y) = x^{2} + y^{2} + x - y - 2$
A) $S (- 2, 1) = 4 + 1 - 2 - 1 - 2 = 0 \Rightarrow$ hence $(- 2, 1)$ lies on circle
B) $S (2, - 1) = 4 + + 2 + 1 - 2 > 0 \Rightarrow$ hence $(2, - 1)$ lies outside circle
C) $S (0, 1) = 0 + 1 + 0 - 1 - 2 < 0 \Rightarrow$ hence $(0, 1)$ lies inside circle
D) $S (2.3) = 4 + 9 + 2 - 3 - 2 > 0$ hence $(2.3)$ lies outside circle
And tangent at $(1, 0)$ is $3 x - y - 3 = 0$ which passes through $(2, 3)$

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