A circle C passes through (2a, 0) and the lin | vedclass

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(C) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$. Let the equation of circle $C$ be $x^2 + y^2 + 2gx + 2fy + c

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centre of $C$ is $(a, 0)$ and $C$ passes through $(0, 0)$ and $(a, a)$

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(C) The radical axis of two circles $S_1 = 0$ and $S_2 = 0$ is given by $S_1 - S_2 = 0$. Let the equation of circle $C$ be $x^2 + y^2 + 2gx + 2fy + c = 0$. The radical axis of $C$ and $x^2 + y^2 - a^2 = 0$ is $(x^2 + y^2 + 2gx + 2fy + c) - (x^2 + y^2 - a^2) = 0$,which simplifies to $2gx + 2fy + c + a^2 = 0$. Given the radical axis is $2x = a$ or $x - a/2 = 0$. Comparing $2gx + 2fy + c + a^2 = 0$ with $x - a/2 = 0$,we get $2g = k$,$2f = 0$,and $c + a^2 = -ak/2$ for some constant $k$. Since $2f = 0$,$f = 0$. Thus,the circle $C$ is $x^2 + y^2 + 2gx + c = 0$. The radical axis is $2gx + c + a^2 = 0$. Comparing with $x = a/2$,we get $2g = 1$ and $c + a^2 = -a/2$. Wait,using the family of circles method: The equation of any circle passing through the intersection of $x^2 + y^2 - a^2 = 0$ and $x - a/2 = 0$ is $(x^2 + y^2 - a^2) + \lambda(x - a/2) = 0$. Since it passes through $(2a, 0)$,we have $(4a^2 - a^2) + \lambda(2a - a/2) = 0$,which gives $3a^2 + \lambda(3a/2) = 0$,so $\lambda = -2a$. The equation of circle $C$ is $x^2 + y^2 - a^2 - 2a(x - a/2) = 0$,which simplifies to $x^2 + y^2 - 2ax = 0$. This is $(x - a)^2 + y^2 = a^2$. The centre is $(a, 0)$. It passes through $(0, 0)$ and $(a, a)$ (since $(a-a)^2 + a^2 = a^2$).

$A$ circle $C$ passes through $(2a, 0)$ and the line $2x = a$ is the radical axis of the circle $C$ and the circle $x^2 + y^2 = a^2$. Then,

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