The Locus of the centroid is a circle with the center centroid of triangle CAB and radius 1/3 radius of a given circle.
Let P be a variable point on the circle with center C, and let CA and CB be perpendicular from C to the x-axis and y-axis, respectively. Let the centroid of triangle PAB be (x,y). We want to show that the locus of (x,y) is a circle with a center at the centroid of triangle CAB and a radius equal to one-third of the radius of the given circle.
The coordinates of point P can be written as (a,b). Then the coordinates of points A and B are (a - c,0) and (0,b - c), respectively.
The centroid of triangle PAB is given by:
((a-c) + 0 + a) / 3, ((0 - c) + (b - c) + b) / 3) = (2a - c) / 3, (2b - c) / 3
We can write these coordinates as:
x = (2a - c) / 3
y = (2b - c) / 3
We can also write the coordinates of the centroid of triangle CAB as:
x0 = (a + c) / 2
y0 = (b + c) / 2
Substituting the values of x and y into the equation of the given circle, we get:
(x - c)^2 + (y - c)^2 = r^2
Substituting the values of x and y into the equation of the circle with center (x0,y0) and radius r/3, we get:
(x - x0)^2 + (y - y0)^2 = (r/3)^2
Simplifying these equations, we get:
2x^2 - 2cx + c^2 + 2y^2 - 2cy + c^2 = r^2
2d^2 - 2cd + c^2 = r^2 / 4
(dc)^2 = (r / 2)^2
This is the equation of a circle with center (c,c/2) and radius r/2. Therefore, the locus of (x,y) is a circle with a center at the centroid of triangle CAB and a radius equal to one-third of the radius of the given circle.