Answer
$(x+1)^2+(y-3)^2=1$
Work Step by Step
With its center at $(-1, 3)$, the tentative equation of the circle is:
$(x-h)^2+(y-k)^2=r^2
\\(x-(-1))^2+(y-3)^2=r^2
\\(x+1)^2+(y-3)^2=r^2$
The circle is tangent to the line $y=2$.
This means that the point on the line $y=2$ that is directly under the center is a point on the circle.
This point is $(-1, 2)$. (refer to the attached image below)
Solve for the value of $r^2$ by substituting the x and y values of the point $(-1, 2)$ into the tentative equation above to obtain:
$(x+1)^2+(y-3)^2=r^2
\\(-1+1)^2+(2-3)^2=r^2
\\0^2+(-1)^2=r^2
\\0+1=r^2
\\1=r^2$
Therefore, the equation of the circle is:
$(x+1)^2+(y-3)^2=1$