Question

give the center and radius of the circle described by the equation and graph the equation. use the graph to identify the domain and range.
$(x + 6)^2 + (y - 3)^2 = 9$
the center is $\square$.
(type an ordered pair. simplify your answer.)
the radius is $\square$.
(type an integer or a simplified fraction.)
graph the circle.
click to enlarge graph
express the domain of the relation in interval notation.
$\square$
express the range of the relation in interval notation.
$\square$

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 and \(r\) is the radius.
For the given equation \((x + 6)^2+(y - 3)^2=9\), we can rewrite it as \((x-(-6))^2+(y - 3)^2=3^2\).

Step2: Identify the center

Comparing with the standard form \((x - h)^2+(y - k)^2=r^2\), we have \(h=-6\) and \(k = 3\). So the center \((h,k)=(-6,3)\).

Step3: Identify the radius

Since \(r^2 = 9\), then \(r=\sqrt{9}=3\).

Step4: Find the domain

The domain represents the set of \(x\) - values. The \(x\) - coordinate of the center is \(h=-6\) and the radius \(r = 3\). The leftmost \(x\) - value is \(h-r=-6-3=-9\) and the rightmost \(x\) - value is \(h + r=-6+3=-3\). So the domain in interval notation is \([-9,-3]\).

Step5: Find the range

The range represents the set of \(y\) - values. The \(y\) - coordinate of the center is \(k = 3\) and the radius \(r=3\). The lowest \(y\) - value is \(k-r=3 - 3=0\) and the highest \(y\) - value is \(k + r=3+3=6\). So the range in interval notation is \([0,6]\).

Answer:

The center is \((-6,3)\).
The radius is \(3\).
The domain is \([-9,-3]\).
The range is \([0,6]\).