Let the center of the circle be $(h,k)$. Since the circle passes through $(0,0)$ and $(0,6)$, the perpendicular bisector of the chord connecting these points must pass through the center. The midpoint of the chord is $(0,3)$. The chord is vertical, so its perpendicular bisector is the horizontal line $y=3$. Thus, $k=3$. Since the center $(h,3)$ lies on the line $2x+3y=15$, substitute $k=3$: $2h+3(3)=15 \Rightarrow 2h+9=15 \Rightarrow 2h=6 \Rightarrow h=3$. So, the center of the circle is $(3,3)$. The radius squared of the circle is $r^2 = (3-0)^2 + (3-0)^2 = 9+9 = 18$. The equation of the circle is $(x-3)^2+(y-3)^2=18$. To find the equation of the tangent at $(0,0)$, we can use the formula $T=0$: $(x-h)(x_1-h) + (y-k)(y_1-k) = r^2$. Substituting $(x_1,y_1)=(0,0)$, $(h,k)=(3,3)$, and $r^2=18$: $(x-3)(0-3) + (y-3)(0-3) = 18 \Rightarrow -3(x-3) -3(y-3) = 18 \Rightarrow -3x+9-3y+9 = 18 \Rightarrow -3x-3y+18=18 \Rightarrow -3x-3y=0$. Dividing by $-3$, we get $x+y=0$. Thus, option C is correct.