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.
$x^{2}+y^{2}-2x - 6y-26 = 0$
the equation in standard form is
(simplify your answer.)

Explanation:

Step1: Group x and y terms

$$(x^{2}-2x)+(y^{2}-6y)=26$$

Step2: Complete the square for x - terms

For \(x^{2}-2x\), using \((a - b)^2=a^{2}-2ab + b^{2}\), here \(a = x\), \(2ab=2x\), so \(b = 1\). Add \(1\) to both sides.
\(x^{2}-2x+1=(x - 1)^{2}\)

Step3: Complete the square for y - terms

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

Step4: Write in standard form

\((x - 1)^{2}+(y - 3)^{2}=26+1 + 9\)
\((x - 1)^{2}+(y - 3)^{2}=36\)

Answer:

The equation in standard form is \((x - 1)^{2}+(y - 3)^{2}=36\)