How are the points $(- 3, - 1), (- 2, 3)$ 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 $x^{2} + y^{2} + 4 x - 2 y + 3 = 0$
Let $A = (- 3, - 1), B = (- 2, 3)$ and $C = (2, 3)$
For point $A (- 3, - 1), x^{2} + y^{2} + 4 x - 2 y + 3 = 9 + 1 - 12 + 2 + 3 = 3 > 0$
Hence point $A$ lies outside the circle
For point $B = (- 2, 3), x^{2} + y^{2} + 4 x - 2 y + 3 = 4 + 9 - 8 - 6 + 3 = 2 > 0$
Hence point $B$ lies outside the circle
For point $C = (2, 3), x^{2} + y^{2} + 4 x - 2 y + 3 = 4 + 9 + 8 - 6 + 3 = 18 > 0$
Hence point $C$ lies outside the circle.

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How are the points (−3,−1),(−2,3) and (2,3) situated with respect to the circle x2+y2+4x–2y+3=0

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