Precalculus (6th Edition)

$(x-2)^2+(y+1)^2=9$ Refer to the graph below.
RECALL: A circle whose equation is of the form $(x-h)^2+(y-k)^2=r^2$ has its center at $(h, k)$ and has a radius of $r$. The given circle has its center $(2, -1)$ and a radius of $3$ units. Substituting these values into the equation above gives: $(x-2)^2+(y-(-1))^2=3^2 \\(x-2)^2+(y+1)^2=9$ Refer to the graph in the answer part above.