$\left\{\begin{array} { l } x=1+y \\ {\left( x-5 \right)}^{2}+{\left( y-4 \right)}^{2}=32\end{array} \right.$
Substitute the given value of $x$ into the equation ${\left( x-5 \right)}^{2}+{\left( y-4 \right)}^{2}=32$${\left( 1+y-5 \right)}^{2}+{\left( y-4 \right)}^{2}=32$
Solve the equation for $y$$\begin{array} { l }y=0,\\y=8\end{array}$
Substitute the given value of $y$ into the equation $x=1+y$$\begin{array} { l }x=1+0,\\y=8\end{array}$
Substitute the given value of $y$ into the equation $x=1+y$$\begin{array} { l }x=1+0,\\x=1+8\end{array}$
Solve the equation for $x$$\begin{array} { l }x=1,\\x=1+8\end{array}$
Solve the equation for $x$$\begin{array} { l }x=1,\\x=9\end{array}$
The possible solutions of the system are the ordered pairs $\left( x, y\right)$$\begin{array} { l }\left( x_1, y_1\right)=\left( 1, 0\right),\\\left( x_2, y_2\right)=\left( 9, 8\right)\end{array}$
Check if the given ordered pairs are the solutions of the system of equations$\begin{array} { l }\left\{\begin{array} { l } 1-0=1 \\ {\left( 1-5 \right)}^{2}+{\left( 0-4 \right)}^{2}=32\end{array} \right.,\\\left\{\begin{array} { l } 9-8=1 \\ {\left( 9-5 \right)}^{2}+{\left( 8-4 \right)}^{2}=32\end{array} \right.\end{array}$
Simplify the equalities$\begin{array} { l }\left\{\begin{array} { l } 1=1 \\ 32=32\end{array} \right.,\\\left\{\begin{array} { l } 9-8=1 \\ {\left( 9-5 \right)}^{2}+{\left( 8-4 \right)}^{2}=32\end{array} \right.\end{array}$
Simplify the equalities$\begin{array} { l }\left\{\begin{array} { l } 1=1 \\ 32=32\end{array} \right.,\\\left\{\begin{array} { l } 1=1 \\ 32=32\end{array} \right.\end{array}$
Since all of the equalities are true, the ordered pairs are the solution of the system$\begin{array} { l }\left( x_1, y_1\right)=\left( 1, 0\right),\\\left( x_2, y_2\right)=\left( 9, 8\right)\end{array}$