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

Answer 1

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.