How do you find the center of the circle given an equation? - Answers

A circle with center (xo, yo) and radius r has an equation of the form: (x - xo)2 + (y - yo)2 = r2 which can be rearranged to: x2 - 2xox + xo2 + y2 - 2yoy + yo2 = r2 ⇒ x2 - 2xox +y2 - 2yoy + xo2 + yo2 - r2 = 0 or x2 - 2xox +y2 - 2yoy = r2 - xo2 - yo2 So depending how your equation is given, the center can be read directly from the equation (first form above) or is -1/2 times the coefficients of the x and y terms. eg: What is the center of the circle with equation x2 - 4x + y2 + 6y + 12 = 0 The coefficients of the x and y terms are -4 and +6, so the center is -1/2 times these, namely; (2, -3). Fully rearranging the circle's equation: x2 - 4x + y2 + 6y + 12 = 0 ⇒ (x - 2)2 - 4 + (y + 3)2 - 9 + 12 = 0 [completing the squares] ⇒ (x - 2)2 + (y + 3)2 - 1 = 0 ⇒ (x - 2)2 + (y + 3)2 = 1 ⇒ circle has centre (2, -3) and radius 1.