Spring Mathematical Tournament

The given system is equivalent to $$|\begin{array}{l}{x}^2-2x+2b-3a=0\\ x+a=y+b\end{array}$$ It has a unique solution $(x_0,y_0)$ if the quadratic equation $x^2-2x+2b-3a=0$ has a unique root $x_0$. This means that $D=1-2b+3a=0$ and $x_0=1$. The condition $x_0^2+y_0^2=1025$ gives $y_0^2=1024$, i.e. $y_0=\pm 2$. For $y_0=-2$ we get $a=x_0^2-2y_0=5$ and $b=x_0+a-y_0=8$, and for $y_0=2$ we get $a=x_0^2-2y_0=-3$ and $b=x_0+a-y_0=-4$.