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. (x - 2)^2+(y - 4)^2 = 16. what is the center of the circle? the circle is centered at the point (type an ordered pair.) what is the radius of the circle? the radius of the circle is (type an integer or a fraction.) use the graphing tool to graph the circle. what is the domain of the relation? the domain is (type your answer in interval notation.)

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. Given the equation $(x - 2)^2+(y - 4)^2 = 16$, we can identify the center and radius.

Step2: Identify the center of the circle

Comparing $(x - 2)^2+(y - 4)^2 = 16$ with $(x - h)^2+(y - k)^2=r^2$, we have $h = 2$ and $k = 4$. So the center of the circle is $(2,4)$.

Step3: Identify the radius of the circle

Since $(x - 2)^2+(y - 4)^2 = 16$ and $r^2=16$, then $r=\sqrt{16}=4$.

Step4: Find the domain of the circle

The $x$ - values of the circle range from $h - r$ to $h + r$. Here, $h = 2$ and $r = 4$. So $x$ ranges from $2-4=-2$ to $2 + 4=6$. In interval notation, the domain is $[-2,6]$.

Answer:

The center of the circle is $(2,4)$. The radius of the circle is $4$. The domain of the relation (the circle) is $[-2,6]$.