Graph the equations to determine whether the system has any solutions. Find any solutions that exist. $$\{\begin{array}{r}{x}^2+y^2=16\\ x-y=-4\end{array}$$

A) $(0,-4),(-4,0)$
B) $(0,-4),(0,4)$
C) $(0,4),(4,0)$
D) $(0,4),(-4,0)$
E) no solution exists

First, let's solve the second equation for y in terms of x:

$$\begin{array}{rl}x-y& =-4\\ y& =x+4\end{array}$$

We can substitute this expression for y into the first equation:

$$\begin{array}{rl}{x}^2+y^2& =16\\ x^2+(x+4{)}^2& =16\\ x^2+x^2+8x+16& =16\\ 2x^2+8x& =0\\ 2x(x+4)& =0\end{array}$$

Solving for x, we get $x=0$ or $x=-4$.

If $x=0$, then $y=4$ or $y=-4$. So one solution is $(0,4)$ and the other is $(0,-4)$.

If $x=-4$, then $y=0$ by the second equation.

Therefore, the system has two solutions: $(0,4)$ and $(-4,0)$. The graph of the two equations is shown below:

\begin{center}
\begin{tikzpicture}[scale=0.7]
\draw[<->,thick] (-5,0)--(5,0);
\draw[<->,thick] (0,-5)--(0,5);
\draw (0,0) circle [radius=4];
\draw[domain=-5:5,smooth,variable=\x,blue] plot ({\x},{\x+4});
\filldraw[black] (0,4) circle (2pt) node[anchor=south west] {$(0,4)$};
\filldraw[black] (0,-4) circle (2pt) node[anchor=north west] {$(0,-4)$};
\filldraw[black] (-4,0) circle (2pt) node[anchor=north east] {$(-4,0)$};
\end{tikzpicture}
\end{center}