How do you find the center and radius of a circle using a polynomial #(x^2) + (y^2) + 6x - 4y = 12#?

1 Answer
Feb 8, 2016

Answer:

Re-write the equation in standard form for a circle to get:
center: #(-3,2)# and radius: #5#

Explanation:

Given
#color(white)("XXX")x^2+y^2+6x-4y=12#

Complete the square for each of the #x# and #y# components:
#color(white)("XXX")(x^2+6xcolor(red)(+9))+(y^2-4ycolor(blue)(+4)) = 12 color(red)(+9)color(blue)(+4)#

Re-write as squared binomials and squared constant:
#color(white)("XXX")(xcolor(green)(+3))^2+(y-color(orange)(2))^2=color(cyan)(5)^2#

Since the standard form for a circle with center #(color(green)(a),color(orange)(b))# and radius #color(cyan)(r)# is
#color(white)("XXX")(x-color(green)(a))^2+(y-color(orange)(b))^2 = color(cyan)(r)^2#

The given circle has a center #(color(green)(-3),color(orange)(2))# and radius #color(cyan)(5)#
graph{x^2+y^2+6x-4y=12 [-11.955, 8.045, -3, 7]}