Question

find the center and the radius of the following circle. x² - 8x + y² + 6y = 0 the center is (type an ordered pair.)

Explanation:

Step1: Complete the square for x-terms

For \(x^2 - 8x\), take half of -8, which is -4, square it: \((-4)^2 = 16\). Add 16 to both sides.
\(x^2 - 8x + 16 + y^2 + 6y = 16\)

Step2: Complete the square for y-terms

For \(y^2 + 6y\), take half of 6, which is 3, square it: \(3^2 = 9\). Add 9 to both sides.
\(x^2 - 8x + 16 + y^2 + 6y + 9 = 16 + 9\)

Step3: Rewrite as perfect squares

\((x - 4)^2 + (y + 3)^2 = 25\)

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.
So, the center \((h, k)\) is \((4, -3)\) and radius \(r = \sqrt{25}=5\). But for the center part:

Answer:

\((4, -3)\)