## Intermediate Algebra for College Students (7th Edition)

$\\(x+3)^2+(y+1)^2=3$
RECALL: The standard form of the equation of a circle with a center of $(h, k)$ and a radius of $r$ units is: $(x-h)^2 + (y-k)^2 = r^2$ The given circle has its center at $(-3, -1)$ and a radius of $\sqrt3$ units. Use the standard form above where $h=-3$, $k=-1$, and $r=\sqrt3$ to obtain: $[x-(-3)]^2 + [y-(-1)]^2 =(\sqrt3)^2 \\(x+3)^2+(y+1)^2=3$