How do you find the center of a circle, given the equation #x^2+y^2+6x-4y+3=0#?
1 Answer
Feb 5, 2016
centre = (-3 , 2)
Explanation:
The general equation of a circle is
# x^2 + y^2 + 2gx + 2fy + c = 0 # this circle has centre = (-g , -f ) and radius r =
# sqrt(g^2+f^2-c) # the equation given here , compared with the general equation will give values of g and f .
from
# x^2 + y^2 + 6x - 4y + 3 = 0 # so 2g = 6 → g = 3 , 2f = -4 → f = -2
hence centre = (-g , -f) = (- 3 , 2 )
