How do you convert the general form of the equation of a circle #2x^2+2y^2+4y=0# to standard form?
1 Answer
Jan 18, 2016
# x^2 + ( y + 1 )^2 = 1 #
Explanation:
The standard form is :
# (x - a )^2 + (y - b )^2 = r^2 # where (a , b ) are the coordinates of centre and r is radius . These have to be found by rearranging the general form .
general form is :
# x^2 + y^2 + 2gx + 2fy + c = 0# the one here :
# 2x^2 + 2y^2 + 4y = 0# (divide both sides by 2 ) :
# x^2 + y^2 + 2y = 0 # ( comparing this to the general form ) : g = 0 , 2f = 2 so f = 1
and c = 0 .centre = ( - g , - f ) = ( 0 ,- 1 ) and
# r = sqrt( g^2 + f^2 - c ) = sqrt(0 + 1^2 - 0 ) = sqrt1 = 1 # equation in standard form is ;
# (x - 0 )^2 + ( y + 1 )^2 = 1^2 #
#rArr x^2 + ( y +1 )^2 = 1#
