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

Answer:

# 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#