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.
$x^2 + y^2 - 4x - 8y - 5 = 0$

Explanation:

Step1: Group the x and y terms

$$(x^{2}-4x)+(y^{2}-8y)=5$$

Step2: Complete the square for the x - terms

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

Step3: Complete the square for the y - terms

For \(y^{2}-8y\), using the formula \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = y\) and \(2ab = 8y\) (so \(b=4\)). Then \(y^{2}-8y=(y - 4)^{2}-16\)

Step4: Substitute back into the equation

\((x - 2)^{2}-4+(y - 4)^{2}-16=5\)

Step5: Simplify to get the standard form

\((x - 2)^{2}+(y - 4)^{2}=5 + 4+16\)
\((x - 2)^{2}+(y - 4)^{2}=25\)

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.

For the equation \((x - 2)^{2}+(y - 4)^{2}=25=(5)^{2}\)

Answer:

• Standard form: \((x - 2)^{2}+(y - 4)^{2}=25\)
• Center: \((2,4)\)
• Radius: \(r = 5\)