How do you graph # x^2+y^2-2x+6y+6=0#?

1 Answer
Feb 10, 2016

Answer:

Re-write the equation in standard circle format;
then draw a circle with the center and radius given by the equation.

Explanation:

Given
#color(white)("XXX")color(red)(x^2)+color(blue)(y^2)color(red)(-2x)color(blue)(+6y)color(green)(+6)=0#

Rearrange grouping the #x# terms, the #y# terms, and with the constant on the right side
#color(white)("XXX")color(red)(x^2-2x)+color(blue)(y^2+6y)=color(green)(-6)#

Complete the square for the #x# sub-expression and the #y# sub-expression:
#color(white)("XXX")color(red)(x^2-2x+1)+color(blue)(y^2+6y+9) = color(green)(-6)color(red)(+1)color(blue)(+9)#

Write as the sum of squared binomials equal to a square
#color(white)("XXX")color(red)((x-1)^2)+color(blue)((y+3)^2=color(green)(2^2)#

Note that the equation for a circle with center #(color(red)(a),color(blue)(b))# and radius #color(green)(r)# is
#color(white)("XXX")(x-color(red)(a))^2+(y-color(blue)(b))^2=color(green)(r)^2#

So we need to draw a circle with center at #(color(red)(1),color(blue)(-3))# and radius #color(green)(2)#

graph{x^2+y^2-2x+6y+6=0 [-5.354, 4.51, -4.946, -0.015]}