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.