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

If a circle passes through the points $(0,0),(0, y)$, then the coordinates of its centre are
  1. $\left(\frac{-x}{2}, \frac{y}{2}\right)$
  2. $\left(\frac{x}{2}, \frac{y}{2}\right)$
  3. $\left(\frac{-x}{2}, \frac{-y}{2}\right)$
  4. $\left(\frac{x}{2}, \frac{-y}{2}\right)$

Solution

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

Asked in: MHT CET 2021 (23 Sep Shift 2)

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