Let a variable point on the ellipse be $P(X,Y)$.

Given ellipse: $\frac{X^2}{4}+\frac{Y^2}{9}=1$

Fixed point is $A(-3,-5)$.

Let the midpoint of $AP$ be $(x,y)$.

Using midpoint formula: $x=\frac{X+(-3)}{2}=\frac{X-3}{2}$ $y=\frac{Y+(-5)}{2}=\frac{Y-5}{2}$

Hence, $X=2x+3, \qquad Y=2y+5$

Substitute into the ellipse equation.

$\frac{(2x+3)^2}{4}+\frac{(2y+5)^2}{9}=1$

Expand: $\frac{4x^2+12x+9}{4}+\frac{4y^2+20y+25}{9}=1$

$x^2+3x+\frac{9}{4}+\frac{4y^2}{9}+\frac{20y}{9}+\frac{25}{9}=1$

Clear denominators by multiplying by $36$.

$36x^2+108x+81+16y^2+80y+100=36$

$36x^2+16y^2+108x+80y+145=0$

Compare with the options.

This matches: $\boxed{36x^2+16y^2+108x+80y+145=0}$

So the correct option is C.

Verification with stored answer:

Stored correct answer = C, which agrees with the derived result.