#### Answer

(a) center-radius form: $\color{blue}{(x+2)^2+(y-2)^2=4}$
(b) general form: $\color{red}{x^2+y^2+4x-4y+4=0}$

#### Work Step by Step

RECALL:
(1) The center-radius form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$ with center at $(h, k)$ and a radius of $r$ units.
(2) The general form of a circle's equation is $x^2+y^2+Dx+Ey+F=0$
The center of the circle can be found by finding the the midpoint of $(-2, 0)$ and $(-2, 4)$. Note that the midpoint of these two points is $(-2, 2)$.
Thus, the circle has its center at $(-2, 2)$.
The radius of the circle can be found by finding the distance from the center $(-2, 0)$ to $(-2, 2)$, which is $2$ units.
Thus, the radius of the circle is $2$.
(a) The circle has its center at $(-2, 2)$ and has a radius of $2$ units.
Therefore, the equation of the circle in center-radius form is:
$(x-(-2))^2+(y-2)^2=2^2
\\\color{blue}{(x+2)^2+(y-2)^2=4}$
(b) To find the equation of the circle in general form, expand the equation in (a) above then put all terms on the left side to obtain:
$(x+2)^2+(y-2)^2=4
\\x^2+4x+4+y^2-4y+4=4
\\x^2+y^2+4x-4y+8=4
\\x^2+y^2+4x-4y+8-4=0
\\\color{red}{x^2+y^2+4x-4y+4=0}$