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.
$x^{2}+y^{2}-4x - 6y-23 = 0$
(simplify your answer.)
the center of the circle is $(2,3)$.
(type an ordered pair.)
the radius of the circle is $r = 6$.
use the graphing tool to graph the circle.

Explanation:

Step1: Group \(x\) and \(y\) terms

$$
LATEXBLOCK0
$$

Step2: Complete the square for \(x\) terms

For \(x^{2}-4x\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=x\), \(2ab = 4x\Rightarrow b = 2\). Add \(4\) to both sides.

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

Step3: Complete the square for \(y\) terms

For \(y^{2}-6y\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=y\), \(2ab=6y\Rightarrow b = 3\). Add \(9\) to both sides.

$$ (x^{2}-4x + 4)+(y^{2}-6y+9)=23+4 + 9 $$

Step4: Write in standard form

$$ (x - 2)^{2}+(y - 3)^{2}=36 $$

Answer:

The standard form of the circle equation is \((x - 2)^{2}+(y - 3)^{2}=36\). The center of the circle is \((2,3)\) and the radius \(r = 6\).