1. Write the hyperbola in standard form

Given: $16x^2-9y^2+32x+36y-164=0$

Group terms and complete squares: $16(x^2+2x)-9(y^2-4y)-164=0$

$16[(x+1)^2-1]-9[(y-2)^2-4]-164=0$

$16(x+1)^2-16-9(y-2)^2+36-164=0$

$16(x+1)^2-9(y-2)^2-144=0$

So, $16(x+1)^2-9(y-2)^2=144$

Divide by $144$: $\frac{(x+1)^2}{9}-\frac{(y-2)^2}{16}=1$

Hence the hyperbola has:

• center $C(-1,2)$
• $a^2=9,\; b^2=16$
• $c^2=a^2+b^2=25 \Rightarrow c=5$

Therefore the foci are: $F_1=(-1-5,2)=(-6,2), \qquad F_2=(-1+5,2)=(4,2)$

2. Take any point $P$ on the hyperbola

Let $P=(x,y)$ be any point on the given hyperbola.

The triangle is formed by the three points $P, F_1, F_2$.

Its centroid $G(X,Y)$ is $X=\frac{x+(-6)+4}{3}=\frac{x-2}{3},$ $Y=\frac{y+2+2}{3}=\frac{y+4}{3}.$

So, $x=3X+2, \qquad y=3Y-4.$

3. Substitute into the hyperbola equation

Original equation: $16x^2-9y^2+32x+36y-164=0$

Substitute $x=3X+2$, $y=3Y-4$:

$16(3X+2)^2-9(3Y-4)^2+32(3X+2)+36(3Y-4)-164=0$

Now expand:

$(3X+2)^2=9X^2+12X+4$ $16(3X+2)^2=144X^2+192X+64$

$(3Y-4)^2=9Y^2-24Y+16$ $-9(3Y-4)^2=-81Y^2+216Y-144$

Also, $32(3X+2)=96X+64$ $36(3Y-4)=108Y-144$

Adding all terms: $144X^2+192X+64-81Y^2+216Y-144+96X+64+108Y-144-164=0$

$144X^2-81Y^2+288X+324Y-324=0$

Divide by $9$: $16X^2-9Y^2+32X+36Y-36=0$

Thus the locus of the centroid is $16x^2-9y^2+32x+36y-36=0$

(where we rename $X,Y$ back to $x,y$).

4. Match with the options

This is exactly:

Option A: $16x^2-9y^2+32x+36y-36=0$

5. Conclusion

The required locus is: $\boxed{16x^2-9y^2+32x+36y-36=0}$

So the correct option is A.