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

a circle

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a parabola

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an ellipse

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a hyperbola

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Solution

The correct option is D a hyperbola
let $C_{1} := x^{2} + y^{2} = a^{2}$
$C_{2} : (x - 2 a)^{2} + y^{2} = 4 a^{2}$
So any point $P$ which moving such that distance from one point is constant distance for the $C_{1} (0, 0) & C_{2} (2 a, 0)$
and $\frac{P C_{2}}{P C_{1}} = \frac{r + 2 a}{r + a} > 1 (where r is radius of the required circle)$
So path will be hyperbola

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