Answer
A circle with center at $C(0,\displaystyle \frac{3}{2})$ and radius $\displaystyle \frac{5}{2}$
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Work Step by Step
Gather terms containing x and y in separate parentheses
$(x^{2})+(y^{2}-3y)=4$
Complete squares:
$ x^{2}\quad$ is already a square
$y^{2}-3y=y^{2}-2\displaystyle \cdot y\cdot\frac{3}{2}+(\frac{3}{2})^{2}-(\frac{3}{2})^{2}=(y-\frac{3}{2})^{2}-\frac{9}{4}$
The equation becomes
$x^{2}+(y-\displaystyle \frac{3}{2})^{2}-\frac{9}{4}=4$
$x^{2}+(y-\displaystyle \frac{3}{2})^{2}=\frac{25}{4}$
A circle with center at $C(0,\displaystyle \frac{3}{2})$ and radius $\displaystyle \frac{5}{2}$
x-intercepts $:\quad$ $\left[\begin{array}{l}
x^{2}+(0-\frac{3}{2})^{2}=\frac{25}{4}\\
x^{2}+\frac{9}{4}=\frac{25}{4}\\
x^{2}=4\\
x=\pm 2 \\
(-2,0),\quad(2,0)
\end{array}\right] $
y-intercepts $:\quad \left[\begin{array}{l}
0+(y-\frac{3}{2})^{2}=\frac{25}{4}\\
y-\frac{3}{2}=\pm\frac{5}{2}\\
y=\frac{3}{2}\pm\frac{5}{2}\\
y=-1,\quad y=4\\
(0,-1),\quad(0,4)
\end{array}\right] $
