If a circle passes through the points \((0,0),(0, y)\), then the coordinates of its centre are

Let \((h, k)\) be the centre of circle
\(\therefore \sqrt{(\mathrm{h}-0)^2+(\mathrm{k}-0)^2}=\sqrt{(\mathrm{h}-\mathrm{x})^2+(\mathrm{k}-0)^2}\) \(=\sqrt{(\mathrm{h}-0)^2+(\mathrm{k}-\mathrm{y})^2} \)
\( \therefore \mathrm{h}^2+\mathrm{k}^2=(\mathrm{h}-\mathrm{x})^2+\mathrm{k}^2 \quad=\mathrm{h}^2+(\mathrm{k}-\mathrm{y})^2 \)
\( \therefore-2 \mathrm{hx}+\mathrm{y}^2=0 \quad \Rightarrow \mathrm{x}(\mathrm{x}-2 \mathrm{~h})=0 \text { and } \)
\( -2 \mathrm{ky}+\mathrm{y}^2=0 \text { and } \mathrm{y}(\mathrm{y}-2 \mathrm{k})=0 \)
\( \therefore \mathrm{x}=0,2 \mathrm{~h} \quad \text { and } \mathrm{y}=0,2 \mathrm{k} \)
\( \therefore \mathrm{x}=2 \mathrm{~h} \text { and } \mathrm{y}=2 \mathrm{k}, \Rightarrow \mathrm{h}=\frac{\mathrm{x}}{2}, \mathrm{k}=\frac{\mathrm{y}}{2}\)