help with this question plz and thankyou

Question

Determine the center and radius of this circle: x^2 +6x +y^2 -8y =0

Answer 1

The "standard form" of a circle is (x - h)² + (y - k)² = r² where the center of the circle is (h, k) and the radius of the circle is "r".

To turn this equation into the "standard form" for a circle, you need to complete the square for both the x-variable and for the y-variable:

x² + 6x + y² - 8y = 0 [Separate the variables:

(x² + 6x ) + (y² - 8y ) = 0 [Now add into each parenthesis the number that will make each into a perfect square. To find that number, take half of the "middle" coefficient (for the "x" that would be 6 and for the "y" that would be "-8") then square that answer and add it in. At the same time, add it to the other side of the equation as well so you don't change the equation:

(x² + 6x + 9) + (y² - 8y + 16) = 0 + 9 + 16 [Now write each quadratic as a square and combine the right side numbers:

(x + 3)² + (y - 4)² = 25 or (x + 3)² + (y - 4)² = 5²

This is the standard form. From it you can read the center of the circle as (-3, 4) and the radius as 5.

Answer 2