Answer
Standard form: $\color{blue}{x^2+(y-2)^2=4}$
General Form $\color{magenta}{x^2+y^2-4y=0}$
Refer to the image below for the graph.
Work Step by Step
RECALL:
The standard form of a circle's equation is:
$(x-h)^2 +(y-k)^2=r^2$
where $r$ = radius and $(h, k)$ is the center.
The circle has:
center: $(h, k)=(0, 2)$
$r=2$
Substitute the given values of $h, k,$ and $r$ into the standard form above to obtain:
$(x-0)^2+(y-2)^2=2^2
\\\color{blue}{x^2+(y-2)^2=4}$
Write the equation in general form by squaring each binomial then subtracting $4$ on both sides of the equation to obtain:
$x^2+(y-2)^2=4
\\x^2+y^2-4y+4=4
\\x^2+y^2-4y+4-4=0
\\\color{magenta}{x^2+y^2-4y=0}$
To graph the circle, perform the following steps:
(1) Plot the center $(0, 2)$.
(2) With a radius of $2$ units, plot the following points:
2 units to the left of the center: $(-2, 2)$
2 units to the right of the center: $(2, 2)$
2 units above the center: $(0, 4)$
2 units below the center: $(0 ,0)$
(3) Connect the four points above (not including the center) using a smooth curve to form a circle
(Refer to the attached image in the answer part above for the graph.)