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.
$(x + 3)^{2}+(y + 5)^{2}=4$
the center is $(-3,-5)$.
(type an ordered pair. simplify your answer.)
the radius is $2$
(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 \((x + 3)^2+(y + 5)^2=4\), we can rewrite it as \((x-(-3))^2+(y - (-5))^2=4\). So, \(h=-3\) and \(k = - 5\). The center \((h,k)=(-3,-5)\).

Step3: Identify the radius

Since \(r^2 = 4\), taking the square root of both sides (and considering \(r>0\) as radius is a non - negative quantity), we get \(r=\sqrt{4}=2\).

Step4: Find the domain

The leftmost \(x\) - value of the circle is \(h - r=-3-2=-5\) and the rightmost \(x\) - value is \(h + r=-3 + 2=-1\). So, the domain is \(\{x|-5\leq x\leq-1\}\).

Step5: Find the range

The lowest \(y\) - value of the circle is \(k - r=-5-2=-7\) and the highest \(y\) - value is \(k + r=-5 + 2=-3\). So, the range is \(\{y|-7\leq y\leq-3\}\).

Answer:

• Center: \((-3,-5)\)
• Radius: \(2\)
• Domain: \(\{x|-5\leq x\leq-1\}\)
• Range: \(\{y|-7\leq y\leq-3\}\)