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(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(2a, 0)$,we have $4a^2 + 4ag + c = 0$ ... $(i)$.The rad

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

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(A) Let the equation of the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$.Since it passes through $(2a, 0)$,we have $4a^2 + 4ag + c = 0$ ... $(i)$.The radical axis of the circle $x^2 + y^2 + 2gx + 2fy + c = 0$ and $x^2 + y^2 - a^2 = 0$ is given by $(2gx + 2fy + c) - (-a^2) = 0$,which simplifies to $2gx + 2fy + c + a^2 = 0$.Given that the radical axis is $x - \frac{a}{2} = 0$,we compare the coefficients:$\frac{2g}{1} = \frac{2f}{0} = \frac{c + a^2}{-a/2}$.From $\frac{2f}{0}$,we get $f = 0$.From $\frac{2g}{1} = \frac{c + a^2}{-a/2}$,we get $c + a^2 = -ag$,or $ag + c + a^2 = 0$ ... $(ii)$.Subtracting $(ii)$ from $(i)$: $(4a^2 + 4ag + c) - (ag + c + a^2) = 0$ $\Rightarrow 3a^2 + 3ag = 0$ $\Rightarrow g = -a$.Substituting $g = -a$ into $(ii)$: $a(-a) + c + a^2 = 0$ $\Rightarrow -a^2 + c + a^2 = 0$ $\Rightarrow c = 0$.Substituting $g = -a, f = 0, c = 0$ into the general equation,we get $x^2 + y^2 - 2ax = 0$.

The equation of the circle which passes through the point $(2a, 0)$ and whose radical axis is $x = \frac{a}{2}$ with respect to the circle $x^2 + y^2 = a^2$ is:

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