Question

A certain circle can be represented by the following equation. x^2+y^2+6y-72=0 What is the center of this circle ?

Answer 1

Answer:

(0, -3)

Step-by-step explanation:

Here we'll rewrite x^2+y^2+6y-72=0 using "completing the square."

Rearranging x^2+y^2+6y-72=0, we get  x^2 + y^2 + 6y                = 72.

x^2 is already a perfect square.  Focus on rewriting y^2 + 6y as the square of a binomial:  y^2 + 6y becomes a perfect square if we add 9 and then subtract 9:

x^2 + y^2 + 6y + 9  - 9              = 72:

x^2 + (y + 3)^2 = 81

Comparing this to the standard equation of a circle with center at (h, k) and radius r,

(x - h)^2 + (y - k)^2 = r^2.  Then h = 0, k = -3 and r = 9.

The center of the circle is (h, k), or (0, -3).