How do you graph #x^2 + y^2 + 6x - 4y - 12 = 0#?
1 Answer
Mar 1, 2016
circle : centre (-3,2) , r = 5
Explanation:
The general form of the equation of a circle is :
#x^2 + y^2 + 2gx + 2fy + c = 0 # centre = (-g,-f) and r
# =sqrt(g^2 + f^2 -c) # the equation
#x^2 + y^2 + 6x - 4y - 12 = 0 " is in this form "# and by comparison: 2g=6 →g=3, 2f=-4 →f=-2 and c = -12
centre = (-3,2) and r=
#sqrt(3^2+(-2)^2+12) = sqrt25 = 5#
