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finding the center of a circle given the equation

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1 Answer

Hello Elisha!

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 ...