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 $(2,4)$
(type an ordered pair.)
the radius of the circle is $r =$

Explanation:

Step1: Recall the standard form of a circle equation

The standard form of a circle equation is \((x - a)^2+(y - b)^2=r^2\), where \((a,b)\) is the center and \(r\) is the radius.

Step2: Extract the radius from the standard form equation

Given the standard form \((x - 2)^2+(y - 4)^2 = 36\). Comparing with \((x - a)^2+(y - b)^2=r^2\), we have \(r^2=36\).

Step3: Solve for \(r\)

Take the square root of both sides: \(r=\sqrt{36}\).

Answer:

\(6\)