$x^{2} + y^{2} + 2 x - 4 y + k = 0$ passes through

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Solution

A) For $k = 0$
$x^{2} + y^{2} + 2 x - 4 y + k = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y + 1 + 4 - 1 - 4 \Rightarrow {(x + 1)}^{2} + {(y - 2)}^{2} = 5$
Point $(0, 0)$ and $(0, 4)$ satisfies this
B) For $k = 1$
$x^{2} + y^{2} + 2 x - 4 y + k = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y + 1 = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y + 1 + 4 - 4 \Rightarrow {(x + 1)}^{2} + {(y - 2)}^{2} = 4$
Point $(1, 2)$ satisfies this
C) For $k = - 15$
$x^{2} + y^{2} + 2 x - 4 y + k = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y - 15 = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y + 1 + 4 - 20 \Rightarrow {(x + 1)}^{2} + {(y - 2)}^{2} = 20$
Point $(3, 4)$ satisfies this
D) For $k = 2$
$x^{2} + y^{2} + 2 x - 4 y + k = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y - 1 = 0 \Rightarrow x^{2} + y^{2} + 2 x - 4 y + 1 + 4 - 6 \Rightarrow {(x + 1)}^{2} + {(y - 2)}^{2} = 2$
Point (0, 1) satisfies

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