The coordinates of the fixed point of the chord cut off by $2x-5y+18=0$ by the circle $ x^2+y^2-6x+2y-54=0$ are

The fixed point of the chord is the midpoint of the intersection points. The circle equation is $(x-3)^2+(y+1)^2=64$, giving center $(3,-1)$ and radius 8. The perpendicular distance from the center to the line $2x-5y+18=0$ is $|2(3)-5(-1)+18|/\sqrt{29}$, which is $29/\sqrt{29}$. The midpoint is the foot of the perpendicular from the center, so we substitute $(x,y) = (3-2k, -1+5k)$ into the chord equation to find $2(3-2k)-5(-1+5k)+18=0$. This simplifies to $29-29k=0$, so $k=1$. Substituting $k=1$ gives the fixed point $(1,4)$.

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