Answer
$(x-2)^2+(y+3)^2=1$
Work Step by Step
The standard equation a circle with radius $r$ and whose center is at $(h,k)$ is:
$(x-h)^2+(y-k)^2=r^2$
Hence, the standard form of the equation of the given circle is
$(x-2)^2+[y-(-3)]^2=1^2\\
(x-2)^2+(y+3)^2=1$
