completing the square for a circle

Sometimes a circle arrives in disguise — written as something like x^2 + y^2 - 6x + 8y - 11 = 0, with the squared terms scattered and no center or radius in sight. Completing the square is the algebraic tidying that folds this expanded mess back into the standard form (x - h)^2 + (y - k)^2 = r^2, where the center and radius are plain to read.

The trick rests on the identity x^2 - 6x = (x - 3)^2 - 9: to turn x^2 - 6x into a perfect square you take half of the x-coefficient (-6/2 = -3), square it (9), and add then subtract it. Work the x-terms and y-terms separately. For x^2 + y^2 - 6x + 8y - 11 = 0: group as (x^2 - 6x) + (y^2 + 8y) = 11; complete each square, (x^2 - 6x + 9) + (y^2 + 8y + 16) = 11 + 9 + 16 — note that whatever you add on the left you must add on the right to keep the equation balanced; this gives (x - 3)^2 + (y + 4)^2 = 36. Now read off: center (3, -4), radius sqrt(36) = 6.

This is the standard way to extract the geometry from a 'general equation of a circle' x^2 + y^2 + Dx + Ey + F = 0. One honest check is built in: after completing the square the right-hand side must be positive to be a genuine circle. If it comes out zero, the 'circle' is a single point (radius 0); if it comes out negative, there are no real points at all — the equation describes an imaginary circle and graphs as nothing. Always read the sign of that final number.