How do you find the center and radius for #x^2+2x+y^2+6y=6#?

1 Answer
Jul 24, 2016

Do a double completion of square, with respect to #x# and #y#, in order to convert to the arm #(x - a)^2 + (y - b)^2 = r#.

#1(x^2 + 2x) + 1(y^2 + 6y) = 6#

#1(x^2 + 2x + 1 - 1) + 1(y^2 + 6y + 9 - 9) = 6#

#1(x^2 + 2x + 1) - 1 + 1(y^2 + 6y + 9) - 9 = 6#

#(x + 1)^2 + (y + 3)^2 - 10 = 6#

#(x + 1)^2 + (y + 3)^2 = 16#

In the form #(x - a)^2 + (y - b)^2 = r#, the radius is given by #sqrt(r)# and the centre is at #(a, b)#. Then, the centre is at #(-1, -3)# and the radius measures #4# units.

Hopefully this helps!