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(D) Let the equation of the circle passing through $(0, 0)$ and $(1, 0)$ be $x^2 + y^2 - x + \lambda y = 0$.This circle has its centre at $\left( \fra

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Both $(A)$ and $(B)$

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(D) Let the equation of the circle passing through $(0, 0)$ and $(1, 0)$ be $x^2 + y^2 - x + \lambda y = 0$.This circle has its centre at $\left( \frac{1}{2}, -\frac{\lambda}{2} \right)$ and radius $r = \sqrt{\left(\frac{1}{2}\right)^2 + \left(\frac{\lambda}{2}\right)^2} = \sqrt{\frac{1 + \lambda^2}{4}} = \frac{\sqrt{1 + \lambda^2}}{2}$.The given circle is $x^2 + y^2 = 9$,which has centre $(0, 0)$ and radius $R = 3$.Since the two circles touch each other,the distance between their centres must be equal to the sum or difference of their radii: $d = |R \pm r|$.The distance between the centres is $d = \sqrt{\left(\frac{1}{2} - 0\right)^2 + \left(-\frac{\lambda}{2} - 0\right)^2} = \sqrt{\frac{1 + \lambda^2}{4}} = \frac{\sqrt{1 + \lambda^2}}{2}$.Thus,$\frac{\sqrt{1 + \lambda^2}}{2} = |3 \pm \frac{\sqrt{1 + \lambda^2}}{2}|$.Case $1$: $\frac{\sqrt{1 + \lambda^2}}{2} = 3 + \frac{\sqrt{1 + \lambda^2}}{2} \Rightarrow 3 = 0$ (Impossible).Case $2$: $\frac{\sqrt{1 + \lambda^2}}{2} = |3 - \frac{\sqrt{1 + \lambda^2}}{2}|$.If $3 - \frac{\sqrt{1 + \lambda^2}}{2} = \frac{\sqrt{1 + \lambda^2}}{2}$,then $\sqrt{1 + \lambda^2} = 3$ $\Rightarrow 1 + \lambda^2 = 9$ $\Rightarrow \lambda^2 = 8$ $\Rightarrow \lambda = \pm 2\sqrt{2}$.If $3 - \frac{\sqrt{1 + \lambda^2}}{2} = -\frac{\sqrt{1 + \lambda^2}}{2}$,then $3 = 0$ (Impossible).Thus,the centres are $\left( \frac{1}{2}, -\frac{\lambda}{2} \right) = \left( \frac{1}{2}, \mp \sqrt{2} \right)$.

The centre$(s)$ of the circle$(s)$ passing through the points $(0, 0)$ and $(1, 0)$ and touching the circle $x^2 + y^2 = 9$ is/are:

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