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

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Solution

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

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

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