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 $\square$.
(simplify your answer.)

Explanation:

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

$$(x^{2}-4x)+(y^{2}-8y)=16$$

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

For \(x^{2}-4x\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2ab=4x\Rightarrow b = 2\). Add \(b^{2}=4\) to both sides.
$$(x^{2}-4x + 4)+(y^{2}-8y)=16+4$$

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

For \(y^{2}-8y\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\), \(2ab = 8y\Rightarrow b=4\). Add \(b^{2}=16\) to both sides.
$$(x^{2}-4x + 4)+(y^{2}-8y+16)=16 + 4+16$$

Step4: Write in standard form

The standard form of a circle is \((x - h)^{2}+(y - k)^{2}=r^{2}\), where \((h,k)\) is the center and \(r\) is the radius.
$$(x - 2)^{2}+(y - 4)^{2}=36$$

Answer:

The equation in standard form is \((x - 2)^{2}+(y - 4)^{2}=36\)