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
(simplify your answer.)

Explanation:

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

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

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

For \(x^{2}-4x\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2ab=4x\Rightarrow b = 2\). Then \(x^{2}-4x=(x - 2)^{2}-4\)

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

For \(y^{2}-6y\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\), \(2ab = 6y\Rightarrow b=3\). Then \(y^{2}-6y=(y - 3)^{2}-9\)

Step4: Substitute back into the equation

$$(x - 2)^{2}-4+(y - 3)^{2}-9=23$$

Step5: Simplify the equation

$$(x - 2)^{2}+(y - 3)^{2}=23 + 4+9$$
$$(x - 2)^{2}+(y - 3)^{2}=36$$

Answer:

The equation in standard form is \((x - 2)^{2}+(y - 3)^{2}=36\)