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A circle passes through $(0,0)$ and $(1,0)$ and touches the circle $x^2+y^2=9$, then the centre of circle is
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The correct answer is:
$\left(\frac{1}{2}, \pm \sqrt{2}\right)$
Radius of the circle $r=\frac{3}{2}$
$\frac{1}{4}+k^2=\frac{9}{4} \Rightarrow k= \pm \sqrt{2}$
$\text { Hence centre is }\left(\frac{1}{2}, \pm \sqrt{2}\right)$
$\frac{1}{4}+k^2=\frac{9}{4} \Rightarrow k= \pm \sqrt{2}$
$\text { Hence centre is }\left(\frac{1}{2}, \pm \sqrt{2}\right)$
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