How do you graph #x^2 + 2x + y^2 + 6y + 6 = 0#?
1 Answer
Feb 10, 2016
circle: centre (-1 , -3 ) , radius = 2
Explanation:
by rewriting the equation as
# x^2+y^2 +2x+6y+6 = 0# and comparing it to the general equation of a circle:
# x^2 +y^2 +2gx + 2fy + c = 0 # by comparison : 2g = 2 →g = 1 , 2f = 6 → f = 3 and c =6
now centre = (-g , -f ) = (-1 , -3 )
and radius
# = sqrt(g^2+f^2 - c ) = sqrt(1^2+3^2-6) =sqrt4 = 2#
