Question

complete the square and write the given equation in standard form. then give the center and radius of the circle and graph the equation.
$x^{2}+y^{2}-4x - 8y - 16 = 0$
the equation in standard form is
$(x - 2)^{2}+(y - 4)^{2}=36$
(simplify your answer.)
the center of the circle is
(type an ordered pair.)

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.

Step2: Identify \(h\) and \(k\) from the given standard - form equation

Given \((x - 2)^{2}+(y - 4)^{2}=36\), by comparing with \((x - h)^{2}+(y - k)^{2}=r^{2}\), we can see that \(h = 2\) and \(k = 4\).

Answer:

\((2,4)\)