How are the points $(- 2, 2), (2, 3)$ 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 $x^{2} + y^{2} + 2 x - 4 y - 3 = 0$
Let $A = (- 2, 2), B = (2, 3)$ and $C = (2, - 3)$
For point $A (- 2, 2), x^{2} + y^{2} + 2 x - 4 y - 3 = 4 + 4 - 4 - 8 - 3 = - 7 < 0$
Hence point $A$ lies inside the circle
For point $B = (2, 3), x^{2} + y^{2} + 2 x - 4 y - 3 = 4 + 9 + 4 - 12 - 3 = - 8 < 0$
Hence point $B$ lies inside the circle
For point $C = (2, - 3), x^{2} + y^{2} + 2 x - 4 y - 3 = 4 + 9 + 4 + 12 - 3 = 26 > 0$
Hence point $C$ lies outside the circle

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

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