The locus of the centre of a circle which alw | vedclass

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

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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 $A(a, 0)$ and $B(-a, 0)$,the distance from the centre to these points must be equal (as they are radii).So,$\sqrt{(h-a)^2 + (k-0)^2} = \sqrt{(h-(-a))^2 + (k-0)^2}$.Squaring both sides,we get $(h-a)^2 + k^2 = (h+a)^2 + k^2$.$h^2 - 2ah + a^2 + k^2 = h^2 + 2ah + a^2 + k^2$.$-2ah = 2ah$,which implies $4ah = 0$,so $h = 0$.The locus of the centre $(h, k)$ is $x = 0$,which is the perpendicular bisector of the line segment joining $(a, 0)$ and $(-a, 0)$.

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

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