Let k be a positive real number and find  k, so that the line is tangent to the circle.
 
\(y=\sqrt{k-x^2}\\ y=-x+3+k\)
 
\(\color{blue}k\notin \{+\mathbb{R}\}\)
 
\(y=\sqrt{k-x^2}\\ y=-x+3+k \)
 
\(y=\sqrt{{\color{blue}1.365}-x^2}\\ y=-x+3\color{blue}-1.365\)
 
\(\color{blue}k\in \{ \approx 1.365,\ \approx-1.365\}\)   determined with graphics. 
 
If +k is replaced by -k in the problem, it is solvable. 
 
 !
 !