1. Write the centroid coordinates

The vertices of the triangle are: $A(a\cos t, a\sin t),\quad B(b\sin t,-b\cos t),\quad C(1,0).$

If the centroid is $(x,y)$, then $x=\frac{a\cos t+b\sin t+1}{3},$ $y=\frac{a\sin t-b\cos t+0}{3}=\frac{a\sin t-b\cos t}{3}.$

So, $3x-1=a\cos t+b\sin t,$ $3y=a\sin t-b\cos t.$

2. Eliminate the parameter $t$

Now square and add the two equations:

$(3x-1)^2+(3y)^2=(a\cos t+b\sin t)^2+(a\sin t-b\cos t)^2.$

Expand the right-hand side:

$=a^2\cos^2 t+b^2\sin^2 t+2ab\sin t\cos t+a^2\sin^2 t+b^2\cos^2 t-2ab\sin t\cos t.$

The cross terms cancel:

$=a^2(\cos^2 t+\sin^2 t)+b^2(\sin^2 t+\cos^2 t).$

Using $\sin^2 t+\cos^2 t=1$,

$(3x-1)^2+(3y)^2=a^2+b^2.$

3. Identify the locus

Hence the locus of the centroid is $\boxed{(3x-1)^2+(3y)^2=a^2+b^2}.$

This matches Option C.

4. Comparison with stored answer

Stored correct answer: C
Derived answer: C

So the derived answer agrees with the stored answer.