Question

determine the center and radius of the circle and sketch the graph. x^2 + y^2 + 8y = 0

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 of the circle and $r$ is the radius. Given $x^{2}+y^{2}+8y = 0$. We know that $y^{2}+8y=(y + 4)^{2}-16$. So the equation becomes $x^{2}+(y + 4)^{2}-16=0$, or $x^{2}+(y + 4)^{2}=16$.

Step2: Identify the center and radius

Comparing $x^{2}+(y + 4)^{2}=16$ with the standard - form $(x - a)^2+(y - b)^2=r^2$, we have $a = 0$, $b=-4$, and $r^{2}=16$, so $r = 4$.

Answer:

Center: $(0,-4)$; Radius: $4$