Let #C=C(h,k)# be the centre of the circle #S# under reference.
Given that, the radius of #S# is #a#, we have,
# S : (x-h)^2+(y-k)^2=a^2#.
Also, #O=O(0,0) in S rArr (0-h)^2+(0-k)^2=a^2#.
#:. h^2+k^2=a^2............................................................(ast)#.
#:. S : (x-h)^2+(y-k)^2=a^2," becomes by virtue of "(ast), #
# S : x^2+y^2-2hx-2ky=0......................................(ast')#.
If #S# intersects #X"-Axis, i.e., y=0"# in #A#, then, from#(ast')#,
# x^2-2hx=0 rArr x=0, or, x=2h#.
#x=0# gives #O(0,0)#, which is readily lies on #S#.
#:. A=A(2h,0)#.
Likewise, #S nn Y"-Axis"={B} rArr B=B(0,2k)#.
So, the vertices of #DeltaOAB# are,
#O(0,0), A(2h,0) and B(0,2k)#.
These give us the centroid #G# of #DeltaOAB# as,
#G((0+2h+0)/3,(0+0+2k)/3)=G(2/3h,2/3k)#.
To get the locus of #G#, we set #G=G(X,Y)# and get,
#X=2/3h, Y=2/3k, or, h=3/2X, k=3/2Y#.
Utilising #(ast)#, we find, #(3/2X)^2+(3/2Y)^2=a^2, or,#
# X^2+Y^2=4/9a^2#.
Changing from #(X,Y)" to conventional "(x,y)#, we have,
# x^2+y^2=4/9a^2#, is the desired locus, which is a circle
with the centre at the origin and radius equaling #2/3#
of the original.
