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(C) The family of circles passing through the intersection of $S_1: x^2 + y^2 - 1 = 0$ and $S_2: x^2 + y^2 - 2x - 4y + 1 = 0$ is given by $S_2 + \lamb

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$x^2 + y^2 - x - 2y = 0$

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(C) The family of circles passing through the intersection of $S_1: x^2 + y^2 - 1 = 0$ and $S_2: x^2 + y^2 - 2x - 4y + 1 = 0$ is given by $S_2 + \lambda S_1 = 0$.$(x^2 + y^2 - 2x - 4y + 1) + \lambda(x^2 + y^2 - 1) = 0$$(1 + \lambda)x^2 + (1 + \lambda)y^2 - 2x - 4y + (1 - \lambda) = 0$Dividing by $(1 + \lambda)$,we get $x^2 + y^2 - \frac{2}{1 + \lambda}x - \frac{4}{1 + \lambda}y + \frac{1 - \lambda}{1 + \lambda} = 0$.The centre is $C = \left( \frac{1}{1 + \lambda}, \frac{2}{1 + \lambda} \right)$ and the radius $r = \sqrt{\left( \frac{1}{1 + \lambda} \right)^2 + \left( \frac{2}{1 + \lambda} \right)^2 - \frac{1 - \lambda}{1 + \lambda}} = \sqrt{\frac{1 + 4 - (1 - \lambda)(1 + \lambda)}{(1 + \lambda)^2}} = \sqrt{\frac{5 - (1 - \lambda^2)}{(1 + \lambda)^2}} = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$.Since the circle touches the line $x + 2y = 0$,the perpendicular distance from the centre to the line equals the radius:$\left| \frac{\frac{1}{1 + \lambda} + 2(\frac{2}{1 + \lambda})}{\sqrt{1^2 + 2^2}} \right| = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$$\left| \frac{5}{\sqrt{5}(1 + \lambda)} \right| = \frac{\sqrt{4 + \lambda^2}}{|1 + \lambda|}$$\sqrt{5} = \sqrt{4 + \lambda^2}$ $\Rightarrow 5 = 4 + \lambda^2$ $\Rightarrow \lambda^2 = 1$ $\Rightarrow \lambda = \pm 1$.For $\lambda = -1$,the equation does not represent a circle. Thus,$\lambda = 1$.Substituting $\lambda = 1$ into the family equation: $2x^2 + 2y^2 - 2x - 4y = 0$,which simplifies to $x^2 + y^2 - x - 2y = 0$.

The equation of the circle passing through the points of intersection of $x^2 + y^2 - 1 = 0$ and $x^2 + y^2 - 2x - 4y + 1 = 0$ and touching the line $x + 2y = 0$ is

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