{х^2+у^2=5 {х-у=м Помогите?))

$\left \{ {{x^2+y^2=5,} \atop {x-y=m;}} \right. \left \{ {{x^2+(x-m)^2=5,} \atop {y=x-m;}} \right. \left \{ {{2x^2-2mx+m^2-5=0,} \atop {y=x-m;}} \right. \\ D_1=(-m)^2-2\cdot(m^2-5)=10-m^2 \geq 0, \\ m^2-10 \leq 0, \\ (m+\sqrt{10})(m-\sqrt{10}) \leq 0, \\ -\sqrt{10} \leq m \leq \sqrt{10}, \\ m\in[-\sqrt{10};\sqrt{10}]; \\ x_{1,2}=\frac{m\pm\sqrt{10-m^2}}{2};$
$\left \{ {{ \left [ {{x=\frac{m-\sqrt{10-m^2}}{2},} \atop {x=\frac{m+\sqrt{10-m^2}}{2};}} \right. } \atop {y=x-m;}} \right. \left [ {{ \left \{ {{x=\frac{m-\sqrt{10-m^2}}{2},} \atop {y=\frac{m-\sqrt{10-m^2}}{2}-m;}} \right. } \atop {\left \{ {{x=\frac{m+\sqrt{10-m^2}}{2},} \atop {y=\frac{m+\sqrt{10-m^2}}{2}-m;}} \right. }} \right.$ $\left [ {{ \left \{ {{x=\frac{m-\sqrt{10-m^2}}{2},} \atop {y=-\frac{m+\sqrt{10-m^2}}{2};}} \right. } \atop {\left \{ {{x=\frac{m+\sqrt{10-m^2}}{2},} \atop {y=\frac{\sqrt{10-m^2}-m}{2};}} \right. }} \right.$