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The four distinct points $(0,0),(2,0),(0,-2)$ and $(k,-2)$ are concyclic, if $k$ is equal to
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Verified Answer
The correct answer is:
$2$
Let the equation of the circle is
$$
\begin{aligned}
& x^2+y^2+2 g x+2 f y+c=0 \\
& \text { at }(0,0), 0+0+0+0+c=0 \Rightarrow c=0 \\
& \text { at } \quad(2,0), 4+0+2 g \cdot 2+0+0=0 \\
& \Rightarrow \quad g=-1 \\
& \text { at }(0,-2), 0+4+0-4 f+0=0 \Rightarrow f=1 \\
& \text { at }(k,-2), k^2+4+2 k g-4 f+0=0 \\
& \Rightarrow \quad k^2+4-2 k-4=0 \\
& \Rightarrow \quad k(k-2)=0 \\
& \Rightarrow \quad k=0, k=2 \\
&
\end{aligned}
$$
$$
\begin{aligned}
& x^2+y^2+2 g x+2 f y+c=0 \\
& \text { at }(0,0), 0+0+0+0+c=0 \Rightarrow c=0 \\
& \text { at } \quad(2,0), 4+0+2 g \cdot 2+0+0=0 \\
& \Rightarrow \quad g=-1 \\
& \text { at }(0,-2), 0+4+0-4 f+0=0 \Rightarrow f=1 \\
& \text { at }(k,-2), k^2+4+2 k g-4 f+0=0 \\
& \Rightarrow \quad k^2+4-2 k-4=0 \\
& \Rightarrow \quad k(k-2)=0 \\
& \Rightarrow \quad k=0, k=2 \\
&
\end{aligned}
$$
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