We are given the following equation of a circle.
[tex]\mleft(x-1\mright)^2+(y+3)^2=4[/tex]The standard form of the equation of a circle is given by
[tex](x-h)^2+(y-k)^2=r^2[/tex]Comparing the given equation with the standard form we see that
[tex]\begin{gathered} h=1 \\ k=-3 \\ r^2=4 \\ r=\sqrt[]{4} \\ r=2 \end{gathered}[/tex]Therefore, the center of the circle is
[tex]C=(h,k)=(1,-3)[/tex]Therefore, the radius of the circle is
[tex]r=2[/tex]