Answer
$(x+1)^2+(y-3)^2=1$
Work Step by Step
Step 1. Given the circle center $(-1,3)$ and tangent to the line $y=2$, which is a horizontal line, we can find the radius as the vertical distance from the center to the line.
The $y$-coordinate of the center of the circle is $3$, so the radius of the circle is the vertical distance from $y=2$ to $y=3$. Thus,
$$\text{radius}=3-2=1$$
Step 2. The standard form of a circle's equation whose center is at $(h, k)$ and radius $r$ is $(h-k)^2+(y-k)^2=r^2$.
Thus, the standard form of the equation of a circle with center at $(-1, 3)$ and a radius of $1$ unit is:
$$\begin{align*}
[x-(-1)])^2+(y-3)^2&=1^2\\
(x+1)^2+(y-3)^2&=1
\end{align*}$$
