Rewrite the equation in the standar form:
[tex](x-h)^2+(y-k)^2=r^2[/tex]So:
[tex]\begin{gathered} x^2+y^2+2x-4y=0 \\ x^2+y^2+2x-4y=(x+1)^2+(y-2)^2-5 \\ so\colon \\ (x+1)^2+(y-2)^2=5 \end{gathered}[/tex]Since the circles are concentric they share the same center, therefore, for the other circle, the Center(h,k) is also:
[tex]C(h,k)=(-1,2)[/tex]Using the standard equation of the circle again:
[tex](x+1)^2+(y-2)^2=r^2[/tex]We need to find the radius, however, we know one of the points of the circle, which is:
[tex](x1,y1)=(5,3)[/tex]Evaluating the point into the equation:
[tex]\begin{gathered} (5+1)^2+(3-2)^2=r^2 \\ 36+1=r^2 \\ 37=r^2 \end{gathered}[/tex]Therefore, the equation is:
[tex](x+1)^2+(y-2)^2=37[/tex]