The given center is (-2,1), which means h = -2 and k = 1. We also know that the circle passes through (-4,1), which means x = -4 and y = 1.
Let's use the standard form first
[tex](x-h)^2+(y-k)^2=r^2[/tex]Let's replace the given information to find r
[tex]\begin{gathered} (-4-(-2))^2+(1-1)^2=r^2 \\ r=\sqrt[]{(-4+2)^2}=\sqrt[]{(-2)^2} \\ r=\sqrt[]{4}=\pm2 \end{gathered}[/tex]Once we have r, we can look for the general form
[tex](x-(-2))^2+(y-1)^2=4[/tex]Then, we solve the binomials
[tex]\begin{gathered} (x+2)^2+(y-1)^2=4 \\ x^2+4x+4+y^2-2y+1=4 \end{gathered}[/tex]At last, we organize all the terms and the left side.
[tex]x^2+y^2+4x-2y+4+1-4=0[/tex]