Answer
$\color{blue}{(x+1)^2+(y-2)^2=4}$
Work Step by Step
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:
$\color{blue}{(x+1)^2+(y-2)^2=4}$
