Question

give the center and radius of the circle described by the equation and graph the equation. use the graph to identify the relations domain and range. x² + y² = 64 what is the center of the circle? the circle is centered at the point (type an ordered pair.)

Explanation:

Step1: Recall the standard - form of a circle equation

The standard - form of a circle equation is $(x - h)^2+(y - k)^2=r^2$, where $(h,k)$ is the center of the circle and $r$ is the radius.
The given equation is $x^{2}+y^{2}=64$, which can be written as $(x - 0)^2+(y - 0)^2 = 8^2$.

Step2: Identify the center

Comparing $(x - 0)^2+(y - 0)^2 = 8^2$ with $(x - h)^2+(y - k)^2=r^2$, we can see that $h = 0$ and $k = 0$. So the center of the circle is $(0,0)$.

Step3: Identify the radius

Since $r^{2}=64$, taking the square - root of both sides (and considering the positive value since $r>0$ for the radius of a circle), we get $r = 8$.

Answer:

The center of the circle is $(0,0)$ and the radius is $8$. The domain of the relation is $[-8,8]$ and the range is $[-8,8]$.