First slide

Definition of a circle

Question

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 has the equation

Moderate

Solution

If the circle $(x - h)^{2} + (y - k)^{2} = r^{2}$ touches both the circle $x^{2} + y^{2} = a^{2}$ and $x^{2} + y^{2} - 4 a x = 0$ externally. Then,
$\sqrt{h^{2} + k^{2}} = r + a$ and $\sqrt{(h - 2 a)^{2} + k^{2}} = r + 2 a$
$\begin{array}{ll} ∴ \sqrt{(h - 2 a)^{2} + k^{2}} - \sqrt{h^{2} + k^{2}} = a \\ \Rightarrow \sqrt{(h - 2 a)^{2} + k^{2}} = a + \sqrt{h^{2} + k^{2}} \\ \Rightarrow (h - 2 a)^{2} + k^{2} = a^{2} + h^{2} + k^{2} + 2 a \sqrt{h^{2} + k^{2}} \\ \Rightarrow - 4 a h + 3 a^{2} = 2 a \sqrt{h^{2} + k^{2}} \\ \Rightarrow (3 a - 4 h)^{2} = 4 (h^{2} + k^{2}) \Rightarrow 12 (h - a)^{2} - 4 k^{2} = 3 a^{2} \end{array}$
Hence, the locus of $(h, k)$ is $12 (x - a)^{2} - 4 y^{2} = 3 a^{2}$

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