Let $c_{1} : x^{2} + y^{2} = 1; C_{2} : (x - 10)^{2} + y^{2} = 1 a n d C_{3}; x^{2} + y^{2} - 10 x - 42 y + 457 = 0$ be three circle.A circle C has been drawn to touch circles $C_{1}$ and $C_{2}$ externally and $C_{3}$ internally. Now circles $C_{1}, C_{2}$ and $C_{3}$ start rolling on the circumference of circle C in anticlockwise direction with constant speed. The centroid of the triangle formed by joining the centres of rolling circles $C_{1}, C_{2}$ and $C_{3}$ lies on

$x^{2} + y^{2} - 12 x - 22 y + 144 = 0$

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$x^{2} + y^{2} - 10 x - 24 y + 144 = 0$

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$x^{2} + y^{2} - 8 x - 20 y + 64 = 0$

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$x^{2} + y^{2} - 4 x - 2 y - 4 = 0$

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Solution

The correct option is B
$x^{2} + y^{2} - 10 x - 24 y + 144 = 0$

The equation of circle C is $(x - 5)^{2} + (y - 12)^{2} = 12^{2}$ This circle also touches x - axis at (5,0) From the geometry, centroid lies on the circle $(x - 5)^{2} + (y - 12)^{2} = 5^{2}$

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Let $c_{1} : x^{2} + y^{2} = 1; C_{2} : (x - 10)^{2} + y^{2} = 1 a n d C_{3}; x^{2} + y^{2} - 10 x - 42 y + 457 = 0$ be three circle.A circle C has been drawn to touch circles $C_{1}$ and $C_{2}$ externally and $C_{3}$ internally. Now circles $C_{1}, C_{2}$ and $C_{3}$ start rolling on the circumference of circle C in anticlockwise direction with constant speed. The centroid of the triangle formed by joining the centres of rolling circles $C_{1}, C_{2}$ and $C_{3}$ lies on