Question

find the center and radius of the circle.
x² + y² + 10y - 39 = 0
the center is .
(type an ordered pair.)

Explanation:

Step1: Complete the square for 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 $x^{2}+y^{2}+10y - 39 = 0$. Since there is no $x$ - linear term, $x^{2}$ is already in the form $(x - 0)^2$. For the $y$ - terms, we complete the square for $y^{2}+10y$. Using the formula $(y + k)^2=y^{2}+2ky + k^{2}$, for $y^{2}+10y$, we have $2k = 10$ (so $k = 5$) and $y^{2}+10y=(y + 5)^2-25$. The equation becomes $x^{2}+(y + 5)^2-25-39=0$.

Step2: Rewrite the equation in standard form

Simplify the equation $x^{2}+(y + 5)^2-25-39 = 0$ to $x^{2}+(y + 5)^2=64$.

Answer:

Center: $(0,-5)$; Radius: $8$