Question

complete the square and write the equation of the circle in standard form. express numbers as integers, simplified fractions,
$x^{2}+y^{2}-2x - 8y + 13 = 0$
the equation in standard form is

Explanation:

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

$$(x^{2}-2x)+(y^{2}-8y)+13 = 0$$

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

For \(x^{2}-2x\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=x\) and \(2ab = 2x\Rightarrow b = 1\). So \(x^{2}-2x=(x - 1)^{2}-1\)

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

For \(y^{2}-8y\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a=y\) and \(2ab=8y\Rightarrow b = 4\). So \(y^{2}-8y=(y - 4)^{2}-16\)

Step4: Substitute back into the equation

$$(x - 1)^{2}-1+(y - 4)^{2}-16+13 = 0$$

Step5: Simplify the equation

$$(x - 1)^{2}+(y - 4)^{2}-4=0$$
$$(x - 1)^{2}+(y - 4)^{2}=4$$

Answer:

\((x - 1)^{2}+(y - 4)^{2}=4\)