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

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through $(0,0)$,we have $c = 0$. Since it passes throug

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$2$

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(D) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$. Since the circle passes through $(0,0)$,we have $c = 0$. Since it passes through $(2,0)$,we have $2^2 + 0^2 + 2g(2) + 2f(0) + 0 = 0$,which gives $4 + 4g = 0$,so $g = -1$. Since it passes through $(0,-2)$,we have $0^2 + (-2)^2 + 2g(0) + 2f(-2) + 0 = 0$,which gives $4 - 4f = 0$,so $f = 1$. The equation of the circle is $x^2 + y^2 - 2x + 2y = 0$. Since the point $(k,-2)$ lies on the circle,we substitute $x = k$ and $y = -2$: $k^2 + (-2)^2 - 2(k) + 2(-2) = 0$ $k^2 + 4 - 2k - 4 = 0$ $k^2 - 2k = 0$ $k(k - 2) = 0$. Since the points must be distinct,$k$ cannot be $0$ (as that would make the point $(0,-2)$,which is already given). Therefore,$k = 2$.

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

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