Center and radius of a circle from the general equation.

Q. Determine the center and radius of a circle whose equation given by $x^2 - 4x + y^2 +10y -71 = 0$.


A. The standard form is given by $(x-h)^2 + (y-k)^2 = r^2$. Hence we should complete the squares for the terms involving $x$ and $y$. For x, the completion of $x^2 - 4x + \_\_$ has a constant term given by $[(-4x/ (2\sqrt{x^2})]^2 = 4$. On the other hand for y, we have $[10y/(2\sqrt{y^2})]^2= 25$. Thus we obtain $$(x-\sqrt{4})^2 +(y- -\sqrt{25})^2 -71 - 29 = 0$$ which we finalize to $(x -2)^2 + (y-(-5))^2 = 100$. Thus we find the center to be (2, -5) and the radius 10.