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What is the radius of the circle passing through the points $(0,0),(a, 0)$ and $(0, b) ?$
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Verified Answer
The correct answer is:
$\frac{1}{2} \sqrt{a^{2}+b^{2}}$
Let $(\mathrm{h}, \mathrm{k})$ be the centre of the circle. Since, circle is passing through $(0,0),(\mathrm{a}, 0)$ and $(0, \mathrm{~b})$, distance between centre and these points would be same and equal to radius. Hence, $h^{2}+k^{2}=(h-a)^{2}+k^{2}=h^{2}+(k-b)^{2}$
$$
\begin{array}{l}
\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{a}^{2}-2 \mathrm{ah}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{b}^{2}-2 \mathrm{bk} \\
\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{a}^{2}-2 \mathrm{ah} \\
\Rightarrow \mathrm{h}=\frac{\mathrm{a}}{2}
\end{array}
$$
Similarly, $\mathrm{k}=\frac{\mathrm{b}}{2}$
$\therefore \quad$ Radius of circle $=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}=\frac{1}{2} \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$
$$
\begin{array}{l}
\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{a}^{2}-2 \mathrm{ah}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{b}^{2}-2 \mathrm{bk} \\
\Rightarrow \mathrm{h}^{2}+\mathrm{k}^{2}=\mathrm{h}^{2}+\mathrm{k}^{2}+\mathrm{a}^{2}-2 \mathrm{ah} \\
\Rightarrow \mathrm{h}=\frac{\mathrm{a}}{2}
\end{array}
$$
Similarly, $\mathrm{k}=\frac{\mathrm{b}}{2}$
$\therefore \quad$ Radius of circle $=\sqrt{\mathrm{h}^{2}+\mathrm{k}^{2}}=\frac{1}{2} \sqrt{\mathrm{a}^{2}+\mathrm{b}^{2}}$
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