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.
$(x + 3)^{2}+(y - 1)^{2}=9$
the radius is 3.
(type an integer or a simplified fraction.)
graph the circle.
express the domain of the relation in interval notation.
$-6,0$
express the range of the relation in interval notation.

Explanation:

Step1: Recall the standard form of a circle's equation

The standard form of a circle's 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 \((x + 3)^2+(y - 1)^2=9\), we can rewrite it as \((x-(-3))^2+(y - 1)^2=3^2\).

Step2: Find the center

Comparing with the standard form \((x - h)^2+(y - k)^2=r^2\), we have \(h=-3\) and \(k = 1\). So the center is \((-3,1)\).

Step3: Find the radius

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

Step4: Find the domain

The domain of a circle \((x - h)^2+(y - k)^2=r^2\) is given by \(h-r\leqslant x\leqslant h + r\).
Substituting \(h=-3\) and \(r = 3\), we get \(-3-3\leqslant x\leqslant-3 + 3\), which is \([-6,0]\).

Step5: Find the range

The range of a circle \((x - h)^2+(y - k)^2=r^2\) is given by \(k-r\leqslant y\leqslant k + r\).
Substituting \(k = 1\) and \(r = 3\), we get \(1-3\leqslant y\leqslant1+3\), which is \([-2,4]\).

Answer:

The center is \((-3,1)\), the radius is \(3\), the domain is \([-6,0]\), and the range is \([-2,4]\).