Answer
$(x+3)^2+(y-1)^2=9$
Work Step by Step
RECALL:
The standard equation of a circle whose center is at the point $(h, k)$ is:
$(x-h)^2+(y-k)^2=r^2$
where $r$ = radius
With its center at $(-3, 1)$, the tentative equation of the given circle is:
$(x-(-3))^2+(y-1)^2=r^2
\\(x+3)^2+(y-1)^2=r^2$
Since the circle is tangent to the $y$-axis, the point on the y-axis that is directly to the right of the center is a point on the circle. This point is $(0, 1)$.
Note that the distance from the center $(-3, 1)$ to the point on the circle $(0, 1)$ is 3 units.
Therefore, the standard form of the equation of the given circle is:
$(x+3)^2+(y-1)^2=3^2
\\(x+3)^2+(y-1)^2=9$
