## College Algebra (10th Edition)

$(x+3)^2+(y-1)^2=9$
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: $(x+3)^2+(y-1)^2=3^2 \\(x+3)^2+(y-1)^2=9$