Locus of centres of the circles which touch the two circles $x^{2} + y^{2} = 16$ and $x^{2} + y^{2} = 16 x$ externally is

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Solution

$x^{2} + y^{2} = 16 \to C_{1} (0, 0), r_{1} = 4$

$r^{2} + y^{2} = 16 x$

$x^{2} + y^{2} - 16 x = 0 \to C_{2} (8, 0), r_{2} = 8$

$r_{r} + 4 = \sqrt{h^{2} + k^{2}}$ ...(1)

$r + 8 = \sqrt{(h + 8)^{2} + k^{2}}$ ...(2)

From eqn. (1) & (2), we get

$r + 8 - r - 4 = \sqrt{(h - 8)^{2} + k^{2}} - \sqrt{h^{2} + k^{2}}$

$4 = \sqrt{(h - 8)^{2} + k^{2}} - \sqrt{h^{2} + k^{2}}$

$4 + \sqrt{h^{2} + k^{2}} = \sqrt{(h - 8)^{2} + k^{2}}$

$16 + (h^{2} + k^{2}) + 8 \sqrt{h^{2} + k^{2}} = h^{2} + 64 - 16 h$

$8 \sqrt{h^{+} k^{2}} = 48 - 16 h$

$\sqrt{h^{2} + k^{2}} 6 - 2 h$

$h^{2} + k^{2} = 36 + 4 h^{2} - 24 h$

$3 h^{2} - k^{2} - 24 h + 36 = 0$

Put $(x, y)$

$3 x^{2} - y^{2} - 24 x + 36 = 0$

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