Example 4 Find the equation of the tangent line to the ellipse $\frac{(x-2)^{2}}{9}+\frac{(y+3)^{2}}{4}=1$ at the known point $P(5,1)$.
Let $x^{\prime}=x-2, y^{\prime}=y+3$. In the new coordinate system $x^{\prime} O^{\prime} y$, the coordinates of point $P$ become $P^{\prime}(3,4)$, and the equation of the ellipse becomes $\frac{x^{\prime 2}}{9}+\frac{y^{\prime 2}}{4}=1$.

Assume the equation of the tangent line passing through point $(3,4)$ is $A x+B y+C=0$. By property 9, we combine the equations $3 A+4 B+C=0$ and $9 A^{2}+4 B^{2}=C^{2}$, and eliminate $C$ to get $B^{2}=-2 A B$. Thus, $B=0$ or $B=-2 A$, leading to $C=3 A$ or $C=5 A$. Therefore, the tangent line equations are $A x-3 A=0$ or $A x-2 A y+5 A=0$, which simplify to $x=3$ and $x-2 y+5$. These are the required equations.