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

12 x^{2} + 12 y^{2} - 92 x + 144 = 0

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12 x^{2} + 4 y^{2} - 96 x + 144 = 0

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x^{2} + y^{2} - 96 x + 144 = 0

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12 x^{2} - 4 y^{2} - 96 x + 144 = 0

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Solution

The correct option is D $12 x^{2} - 4 y^{2} - 96 x + 144 = 0$
Given,
Circle $C_{1} : x^{2} + y^{2} = 16$
Center $(0, 0)$ and radius= $4$

Circle $C_{2} : x^{2} + y^{2} = 16 x$
center $(8, 0)$ and radius $= 8$

Let $r$ be the radius and $(h, k)$ be the center of required circle

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

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

Subtract eqs.(1) from (1)-

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

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

Squaring on Both sides

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

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

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

$\sqrt{h^{2} + k^{2}} = 6 - 2 h$
Squaring on both sides
$(h^{2} + k^{2}) = 36 + 4 h^{2} - 24 h ⟹ 3 h^{2} - k^{2} - 24 h + 36 = 0$

Therefore the locus is
$4 (3 x^{2} - y^{2} - 24 x + 36 = 0)$
$12 x^{2} - 4 y^{2} - 96 x + 144 = 0$

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Locus of centres of the circles which touche the two circles x2+y2=16 and x2+y2=16x externally is

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