The locus of the centres of the circles which | vedclass

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(A) Let the centre of the circle be $(h, k)$ and its radius be $r$.Since the circle touches $x^2 + y^2 = a^2$ externally,the distance between their ce

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$12x^2 - 4y^2 - 24ax + 9a^2 = 0$

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(A) Let the centre of the circle be $(h, k)$ and its radius be $r$.Since the circle touches $x^2 + y^2 = a^2$ externally,the distance between their centres is equal to the sum of their radii:$\sqrt{h^2 + k^2} = r + a \implies r = \sqrt{h^2 + k^2} - a$  $(i)$Since the circle touches $x^2 + y^2 - 4ax = 0$ (which is $(x-2a)^2 + y^2 = (2a)^2$) externally,the distance between their centres is:$\sqrt{(h-2a)^2 + k^2} = r + 2a$  $(ii)$Substituting $r$ from $(i)$ into $(ii)$:$\sqrt{(h-2a)^2 + k^2} = (\sqrt{h^2 + k^2} - a) + 2a$$\sqrt{(h-2a)^2 + k^2} = \sqrt{h^2 + k^2} + a$Squaring both sides:$(h-2a)^2 + k^2 = (h^2 + k^2) + a^2 + 2a\sqrt{h^2 + k^2}$$h^2 - 4ah + 4a^2 + k^2 = h^2 + k^2 + a^2 + 2a\sqrt{h^2 + k^2}$$3a^2 - 4ah = 2a\sqrt{h^2 + k^2}$Dividing by $a$ (assuming $a 
eq 0$):$3a - 4h = 2\sqrt{h^2 + k^2}$Squaring again:$(3a - 4h)^2 = 4(h^2 + k^2)$$9a^2 - 24ah + 16h^2 = 4h^2 + 4k^2$$12h^2 - 4k^2 - 24ah + 9a^2 = 0$Replacing $(h, k)$ with $(x, y)$,the locus is $12x^2 - 4y^2 - 24ax + 9a^2 = 0$.

The locus of the centres of the circles which touch externally the circles $x^2 + y^2 = a^2$ and $x^2 + y^2 - 4ax = 0$ will be

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