How are the points $(- 1, 1), (- 2, 2)$ and $(- 2, 3)$ situated with respect to the circle $x^{2} + y^{2} - 4 x - 2 y + 3 = 0$

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Solution

Given equation is $x^{2} + y^{2} + 2 x - 4 y - 3 = 0$
Let $A = (- 1, 1), B = (- 2, 2)$ and $C = (- 2, 3)$
For point $A (- 1, 1), x^{2} + y^{2} + 2 x - 4 y - 3 = 1 + 1 - 2 - 4 - 3 = - 7 < 0$
Hence point $A$ lies inside the circle
For point $B = (- 2, 2), x^{2} + y^{2} + 2 x - 4 y - 3 = 4 + 4 - 4 + 8 - 3 = 9 > 0$
Hence point $B$ lies outside the circle
For point $C = (- 2, 3), x^{2} + y^{2} + 2 x - 4 y - 3 = 4 + 9 - 4 - 12 - 3 = - 6 < 0$
Hence point $C$ lies inside the circle

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