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 center is $(-3,1)$
(type an ordered pair. simplify your answer.)
the radius is 3
(type an integer or a simplified fraction.)
graph the circle.

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.

Step2: Identify the center

For the equation \((x + 3)^2+(y - 1)^2=9\), we can rewrite it as \((x-(-3))^2+(y - 1)^2=3^2\). Comparing with the standard form \((x - h)^2+(y - k)^2=r^2\), we get \(h=-3\) and \(k = 1\). So the center \((h,k)=(-3,1)\).

Step3: Identify the radius

Since the equation is \((x-(-3))^2+(y - 1)^2=3^2\), comparing with \((x - h)^2+(y - k)^2=r^2\), we have \(r = 3\).

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 simplifies to \(-6\leqslant x\leqslant0\).

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 simplifies to \(-2\leqslant y\leqslant4\).

Answer:

Center: \((-3,1)\)
Radius: \(3\)
Domain: \([-6,0]\)
Range: \([-2,4]\)