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.
$(x + 3)^{2}+(y + 5)^{2}=4$
(type an ordered pair. simplify your answer.)
the radius is 2.
(type an integer or a simplified fraction.)
graph the circle.
express the domain of the relation in interval notation.
$-5,-1$
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 \((x + 3)^2+(y + 5)^2=4\), we can rewrite it as \((x-(-3))^2+(y-(-5))^2 = 2^2\).

Step2: Find the center

Comparing with the standard form, \(h=-3\) and \(k = -5\). So the center is \((-3,-5)\).

Step3: Find the range

The range of a circle \((x - h)^2+(y - k)^2=r^2\) is \([k - r,k + r]\).
Here \(k=-5\) and \(r = 2\). So \(k-r=-5-2=-7\) and \(k + r=-5 + 2=-3\).

Answer:

The center is \((-3,-5)\).
The range is \([-7,-3]\)