The four distinct points (0, 0), (2, 0), (0, | vedclass

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(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(0, 0)$,we get $c = 0$.Since it passes through $(2, 0

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$2$

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(B) The general equation of a circle is $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(0, 0)$,we get $c = 0$.Since it passes through $(2, 0)$,we get $4 + 4g = 0 \Rightarrow g = -1$.Since it passes through $(0, -2)$,we get $4 - 4f = 0 \Rightarrow f = 1$.The equation of the circle is $x^2 + y^2 - 2x + 2y = 0$.For the point $(k, -2)$ to be on the circle,it must satisfy the equation:$k^2 + (-2)^2 - 2(k) + 2(-2) = 0$$k^2 + 4 - 2k - 4 = 0$$k^2 - 2k = 0$$k(k - 2) = 0$Thus,$k = 0$ or $k = 2$.Since the points must be distinct,$(k, -2)$ cannot be $(0, -2)$,so $k 
eq 0$.Therefore,$k = 2$.

The four distinct points $(0, 0), (2, 0), (0, -2)$ and $(k, -2)$ are concyclic,if $k =$

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