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.
$(x + 6)^{2}+(y - 1)^{2}=36$
the center is $(-6,1)$
(type an ordered pair. simplify your answer.)
the radius is 6
(type an integer or a simplified fraction.)
graph the circle.
click toenlargegraph

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 \((x + 6)^{2}+(y - 1)^{2}=36\), we can rewrite \((x + 6)\) as \((x-(-6))\). Comparing with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we have \(h=-6\) and \(k = 1\). So the center \((h,k)=(-6,1)\).

Step3: Identify the radius

Since the equation is \((x + 6)^{2}+(y - 1)^{2}=36\), and \(r^{2}=36\), then \(r=\sqrt{36}=6\) (we take the positive value since radius is non - negative).

Step4: Find the domain

The leftmost \(x\) - value is \(h - r=-6-6=-12\) and the rightmost \(x\) - value is \(h + r=-6 + 6=0\). So the domain is \(\{x|-12\leq x\leq0\}\) (or in interval notation \([-12,0]\)).

Step5: Find the range

The lowest \(y\) - value is \(k - r=1-6=-5\) and the highest \(y\) - value is \(k + r=1 + 6=7\). So the range is \(\{y|-5\leq y\leq7\}\) (or in interval notation \([-5,7]\)).

Answer:

Center: \((-6,1)\)
Radius: \(6\)
Domain: \([-12,0]\)
Range: \([-5,7]\)