Question

find the center and radius of the circle

$x^{2}+y^{2}+6y - 7 = 0$

the center is
(type an ordered pair)

the radius is
(simplify your answer. type an exact answer, using radicals as needed)

Explanation:

Step1: Rewrite the equation in standard form.

Complete the square for the $y$ - terms. The general equation of a circle is $(x - a)^2+(y - b)^2=r^2$, where $(a,b)$ is the center and $r$ is the radius.
Given $x^{2}+y^{2}+6y - 7=0$.
We have $x^{2}+(y^{2}+6y)=7$.
To complete the square for $y^{2}+6y$, we take half of the coefficient of $y$ (which is $6$), square it. Half of $6$ is $3$, and $3^{2}=9$. So we add $9$ to both sides of the equation: $x^{2}+y^{2}+6y + 9=7 + 9$.
This gives us $x^{2}+(y + 3)^{2}=16$.

Step2: Identify the center and radius.

Comparing $x^{2}+(y + 3)^{2}=16$ with $(x - a)^2+(y - b)^2=r^2$, we have $a = 0$, $b=-3$ and $r^{2}=16$.
Since $r^{2}=16$, then $r = 4$.

Answer:

The center is $(0,-3)$
The radius is $4$