Answer
Standard Form: $x^2 + (y - 2)^2 = 4$
General Form: $x^2 + y^2 - 4y = 0$.
Refer to the graph below.
Work Step by Step
The standard form of acircle's equation is $(x-h)^2+(y-k)^2=r^2$ where $r$=radius and $(h, k)$ is the center.
Substitute $r=2$, $h=0$ and $k=2$ to both $h$ and $k$ in the standard form above to obtain:
\begin{align*}
(x-0)^2+(y-2)^2&=2^2\\
x^2+(y-2)^2&=4
\end{align*}
General form of the equation of a circle is $x^2+y^2 + ax+by+c=0$.
Subtract $4$ from each side of the equation above to obtain:
\begin{align*}
x^2+(y-2)^2-4&=4-4\\
x^2+y^2-4y+4-4&=0\\
x^2+y^2-4y&=0
\end{align*}
With $r=2$ and center at $(0, 2)$, plot the points $2$ units directly above, below, to the left, and to the right of the circle$. Then, connect these four points using a smooth curve to form a circle.
Refer to the graph above.
