Answer
$(x+3)^2+(y+2)^2=1$
Work Step by Step
RECALL:
The standard form of the equation of a circle with a center at $(h, k)$ and a radius of $r$ units is:
$(x-h)^2+(y-k)^2=r^2$
The given circle has its center at $(-3, -2)$.
The point on the circle that is directly to the left of the center is $(-4, -2)$.
This point is 1 unit away from the center.
This means that the radius is 1 unit.
Since the center is at (-3, -2), we know that h=-3 and k = -2.
The radius is 1 unit, so r = 1.
Therefore, the equation of the given circle is:
$[(x-(-3)]^2 + [(y-(-2)]^2=1^2
\\(x+3)^2+(y+2)^2=1$