How to find the center and radius of #2x^2 + 2y^2 -8x +12y +8=0#?
1 Answer
Jan 29, 2016
centre = (2 , -3 ) and r = 3
Explanation:
The general equation of a circle is
# : x^2 + y^2 + 2gx + 2fy + c = 0# For the given equation to be compared require to divide by 2 .
hence equation is
# x^2 + y^2 - 4x + 6y + 4 = 0# Comparing the equation to the general form.
then 2g = - 4 → g = - 2 and 2f = 6 →f = 3 , c=4
centre = ( - g , - f ) = (2 , - 3 )
and r =
# sqrt(g^2 + f^2 - c ) = sqrt((-2)^2 + 3^2 - 4 ) = 3#
