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

1 Answer
Steve M
Nov 24, 2016

# x^2+2x+y^2+6y+6=0 #

Gather up the terms in #x#, and the terms in #y# and complete the square by combing half the #x# and #y# coefficient for both as follows;

# :. (x+2/2)^2-(2/2)^2 +(y+6/2)^2 - (6/2)^2 +6=0 #
# :. (x+1)^2-(1)^2 +(y+3)^2 - (3)^2 +6=0 #
# :. (x+1)^2-1 +(y+3)^2 - 9 +6=0 #
# :. (x+1)^2 +(y+3)^2 =4 #
# :. (x+1)^2 +(y+3)^2 =2^2 #

A circle of centre #(a,b)# and radius #r# has equation

# (x-a)^2 +(y-b)^2 =r^2 #

Comparing with the above equation we can see that

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

represents a circle of centre (-1,-3) and radius 2

graph{(x+1)^2 +(y+3)^2 =2^2 [-11.17, 8.83, -7.3, 2.7]}

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