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.
$(x + 5)^{2}+(y - 4)^{2}=16$
the center is
(type an ordered pair. simplify your answer.)
the radius is
(type an integer or a simplified fraction.)
graph the circle.
click to enlarge graph
express the domain of the relation in interval notation.
express the range of the relation 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 and \(r\) is the radius.
For the given equation \((x + 5)^2+(y - 4)^2=16\), we can rewrite \((x + 5)\) as \((x-(- 5))\).

Step2: Find the center

Comparing \((x-(-5))^2+(y - 4)^2=16\) with \((x - h)^2+(y - k)^2=r^2\), we get \(h=-5\) and \(k = 4\). So the center \((h,k)=(-5,4)\).

Step3: Find the radius

Since \(r^2 = 16\), then \(r=\sqrt{16}=4\) (we take the positive value as radius is non - negative).

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 \(h=-5\) and \(r = 4\), we have \(-5-4\leqslant x\leqslant-5 + 4\), i.e., \(-9\leqslant x\leqslant-1\). In interval notation, the domain is \([-9,-1]\).

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 \(k = 4\) and \(r = 4\), we have \(4-4\leqslant y\leqslant4 + 4\), i.e., \(0\leqslant y\leqslant8\). In interval notation, the range is \([0,8]\).

Answer:

The center is \((-5,4)\).
The radius is \(4\).
The domain is \([-9,-1]\).
The range is \([0,8]\).