## College Algebra (10th Edition)

$\color{blue}{(x+1)^2+(y-2)^2=4}$
The given circle has its center at $(-1, 2)$. RECALL: The standard form of a circle's equation is $(x-h)^2+(y-k)^2=r^2$, where $(h, k)$ is the center and $r$ is the radius. Thus, using the standard form above with the center of $(-1, 2)$, the tentative equation of the circle whose graph is given is: $(x-(-1))^2 + (y-2)^2=r^2 \\(x+1)^2+(y-2)^2=r^2$ The point $(-1, 0)$ is a point on the circle. This means that the x and y coordinates of this point satisfy the equation of the circle. Substitute the x and y coordinates of this point into the tentative equation above to obtain: $(x+1)^2+(y-2)^2=r^2 \\(-1+1)^2+(0-2)^2=r^2 \\0^2 + (-2)^2=r^2 \\0+4=r^2 \\4=r^2$ Therefore, the equation of the circle is: $\color{blue}{(x+1)^2+(y-2)^2=4}$