what is the center of the circle with the equation: x squared plus y squared minus 6x plus 2y minus 6 equals 0
finding the center of a circle given the equation
A unit circle centered at the origin (0,0) has the form (x-0)^2 + (y-0)^2 = 1
To get -6x, we need (x-3)^2 = x^2 - 6x + 9
To get +2y, we need (y+1)^2 = y^2 + 2y + 1
Combining, we have x^2 + y^2 - 6x + 2y + 10 and set this equal to 16, which is the radius squared.
So, (x-3)^2 + (y+1)^2 = 16 matches x^2 + y^2 - 6x + 2y + 10 = 16,
giving x^2 + y^2 - 6x + 2y - 6 = 0
Circle center is (3,-1) and the radius is 4 units.
Check top point (3,3) that's 4 up from center; check left point (-1,-1) that's 4 left of center. Regards ...