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If a circle passes through the points $(0,0),(0, y)$, then the coordinates of its centre are
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Verified Answer
The correct answer is:
$\left(\frac{x}{2}, \frac{y}{2}\right)$
Let $(h, k)$ be the centre of circle
$$
\begin{aligned}
& \therefore \sqrt{(\mathrm{h}-0)^2+(\mathrm{k}-0)^2}=\sqrt{(\mathrm{h}-\mathrm{x})^2+(\mathrm{k}-0)^2}=\sqrt{(\mathrm{h}-0)^2+(\mathrm{k}-\mathrm{y})^2} \\
& \therefore \mathrm{h}^2+\mathrm{k}^2=(\mathrm{h}-\mathrm{x})^2+\mathrm{k}^2 \quad=\mathrm{h}^2+(\mathrm{k}-\mathrm{y})^2 \\
& \therefore-2 \mathrm{hx}+\mathrm{y}^2=0 \quad \Rightarrow \mathrm{x}(\mathrm{x}-2 \mathrm{~h})=0 \text { and } \\
& -2 \mathrm{ky}+\mathrm{y}^2=0 \text { and } \mathrm{y}(\mathrm{y}-2 \mathrm{k})=0 \\
& \therefore \mathrm{x}=0,2 \mathrm{~h} \quad \text { and } \mathrm{y}=0,2 \mathrm{k} \\
& \therefore \mathrm{x}=2 \mathrm{~h} \text { and } \mathrm{y}=2 \mathrm{k}, \Rightarrow \mathrm{h}=\frac{\mathrm{x}}{2}, \mathrm{k}=\frac{\mathrm{y}}{2}
\end{aligned}
$$
$$
\begin{aligned}
& \therefore \sqrt{(\mathrm{h}-0)^2+(\mathrm{k}-0)^2}=\sqrt{(\mathrm{h}-\mathrm{x})^2+(\mathrm{k}-0)^2}=\sqrt{(\mathrm{h}-0)^2+(\mathrm{k}-\mathrm{y})^2} \\
& \therefore \mathrm{h}^2+\mathrm{k}^2=(\mathrm{h}-\mathrm{x})^2+\mathrm{k}^2 \quad=\mathrm{h}^2+(\mathrm{k}-\mathrm{y})^2 \\
& \therefore-2 \mathrm{hx}+\mathrm{y}^2=0 \quad \Rightarrow \mathrm{x}(\mathrm{x}-2 \mathrm{~h})=0 \text { and } \\
& -2 \mathrm{ky}+\mathrm{y}^2=0 \text { and } \mathrm{y}(\mathrm{y}-2 \mathrm{k})=0 \\
& \therefore \mathrm{x}=0,2 \mathrm{~h} \quad \text { and } \mathrm{y}=0,2 \mathrm{k} \\
& \therefore \mathrm{x}=2 \mathrm{~h} \text { and } \mathrm{y}=2 \mathrm{k}, \Rightarrow \mathrm{h}=\frac{\mathrm{x}}{2}, \mathrm{k}=\frac{\mathrm{y}}{2}
\end{aligned}
$$
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