## College Algebra (10th Edition)

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.
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.)