Question

write the standard form of the equation of the circle described below. center (0, - 3) passes through the point (- 3, - 1)
the standard form of the equation of the circle is
(type an equation. simplify your answer )

Explanation:

Step1: Recall circle - equation formula

The standard form of the equation of a circle is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius. Here, $h = 0$ and $k=-3$, so the equation is $(x - 0)^2+(y + 3)^2=r^2$, or $x^{2}+(y + 3)^{2}=r^{2}$.

Step2: Calculate the radius

The circle passes through the point $(-3,-1)$. Substitute $x=-3$ and $y = - 1$ into the equation $x^{2}+(y + 3)^{2}=r^{2}$. We get $(-3)^{2}+(-1 + 3)^{2}=r^{2}$. Then $9+4=r^{2}$, so $r^{2}=13$.

Answer:

$x^{2}+(y + 3)^{2}=13$