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.
$x^{2}+y^{2}-2x - 6y-26 = 0$
the equation in standard form is $(x - 1)^{2}+(y - 3)^{2}=36$
(simplify your answer.)
the center of the circle is $(1,3)$
(type an ordered pair.)
the radius of the circle is $r = square$.

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 \(r^2\) from the given standard - form equation

We have the equation \((x - 1)^2+(y - 3)^2 = 36\). Comparing with \((x - h)^2+(y - k)^2=r^2\), we know that \(r^2=36\).

Step3: Solve for \(r\)

Take the square root of both sides of the equation \(r^2 = 36\). Since \(r>0\) (radius is a non - negative quantity), \(r=\sqrt{36}\).

Answer:

\(6\)