Let C1:x2+y2=1;C2:(x−10)2+y2=1and C3:x2+y2−10x−42y+457=0 be three circles. A circle C has been drawn to touch circles C1 and C2 externally and C3 internally. Now circles C1, C2 and C3 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 C1, C2 and C3 lies on

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Let $C_{1} : x^{2} + y^{2} = 1; C_{2} : {(x - 10)}^{2} + y^{2} = 1$ and $C_{3} : x^{2} + y^{2} - 10 x - 42 y + 457 = 0$ be three circles. A circle C has been drawn to touch circles C₁ and C₂ externally and C₃ internally. Now circles C₁, C₂ and C₃ 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 C₁, C₂ and C₃ lies on

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

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

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

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

answer is B.

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Detailed Solution

The equation of circle C is ${(x - 5)}^{2} + {(y - 12)}^{2} = 12^{2}$
This circle also touches x–axis at (5, 0).
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From the geometry, centroid lies on the circle(x – 5)² + (y – 12)² = 5².