Question

Let $C_1:x^2+y^2=1;C_2:(x-10{)}^2+y^2=1$ and $C_3:x^2+y^2-10x–42y+457=0$ be three circles. 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

• A. $x^2+y^2-12x-2y=0$
• B. $x^2+y^2-10x-24y+144=0$
• C. $x^2+y^2-8x-20y+64=0$
• D. $x^2+y^2-4x-2y-40=0$

Answer 1

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

The equation of circle C is
$(x-5{)}^2+(y-12{)}^2=122$
This circle also touches x-axis at (5, 0).
From the geometry, centroid lies on the circle $(x-5{)}^2+(y-12{)}^2=52$.