The locus of the centre of the circles which | vedclass

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(D) Let the centre of the variable circle be $(h, k)$ and its radius be $r$.Since the variable circle touches the circle $x^{2}+y^{2}=a^{2}$ externall

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a hyperbola

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(D) Let the centre of the variable circle be $(h, k)$ and its radius be $r$.Since the variable circle touches the circle $x^{2}+y^{2}=a^{2}$ externally, the distance between their centres is equal to the sum of their radii:$\sqrt{h^{2}+k^{2}} = r + a \implies h^{2}+k^{2} = (r+a)^{2} \quad (1)$Since it also touches the circle $x^{2}+y^{2}=4ax$ (which has centre $(2a, 0)$ and radius $2a$) externally:$\sqrt{(h-2a)^{2}+k^{2}} = r + 2a \implies (h-2a)^{2}+k^{2} = (r+2a)^{2} \quad (2)$Subtracting equation $(1)$ from equation $(2)$:$(h-2a)^{2} - h^{2} = (r+2a)^{2} - (r+a)^{2}$$h^{2} - 4ah + 4a^{2} - h^{2} = r^{2} + 4ar + 4a^{2} - (r^{2} + 2ar + a^{2})$$-4ah + 4a^{2} = 2ar + 3a^{2}$$2ar = a^{2} - 4ah \implies r = \frac{a-4h}{2}$Substituting $r$ back into equation $(1)$:$h^{2}+k^{2} = (\frac{a-4h}{2} + a)^{2} = (\frac{3a-4h}{2})^{2}$$4(h^{2}+k^{2}) = 9a^{2} - 24ah + 16h^{2}$$12h^{2} - 4k^{2} - 24ah + 9a^{2} = 0$Replacing $(h, k)$ with $(x, y)$, the locus is $12x^{2}-4y^{2}-24ax+9a^{2}=0$, which represents a hyperbola.

The locus of the centre of the circles which touch both the circles $x^{2}+y^{2}=a^{2}$ and $x^{2}+y^{2}=4ax$ externally is

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