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(D) Let the centre of the circle be $(h, k)$.Since the circle passes through the points $(a, 0)$ and $(-a, 0)$,the distance from the centre $(h, k)$ t

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$x=0$

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(D) Let the centre of the circle be $(h, k)$.Since the circle passes through the points $(a, 0)$ and $(-a, 0)$,the distance from the centre $(h, k)$ to both points must be equal to the radius $r$.Therefore,$\sqrt{(h-a)^2 + (k-0)^2} = \sqrt{(h-(-a))^2 + (k-0)^2}$.Squaring both sides,we get $(h-a)^2 + k^2 = (h+a)^2 + k^2$.Simplifying this,$h^2 - 2ah + a^2 + k^2 = h^2 + 2ah + a^2 + k^2$.This leads to $-2ah = 2ah$,which implies $4ah = 0$.Since $a 
eq 0$,we must have $h = 0$.Replacing $(h, k)$ with $(x, y)$,the locus of the centre is $x = 0$,which is the $y$-axis.

The locus of the centre of a circle which passes through two fixed points $(a, 0)$ and $(-a, 0)$ is

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