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.

$x^{2}+y^{2}-4x - 6y-23 = 0$

the equation in standard form is $(x - 2)^{2}+(y - 3)^{2}=36$.
(simplify your answer.)

the center of the circle is $\square$.
(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 \((x - 2)^2+(y - 3)^2 = 36\), by comparing with \((x - h)^2+(y - k)^2=r^2\), we have \(h = 2\) and \(k=3\).

Answer:

\((2,3)\)