In the general form #2x^2 + 2y^2 + 4x -8y -22 = 0#, how do you convert to standard form?
1 Answer
Feb 6, 2016
# (x+1)^2 + (y-2 )^2 = 16#
Explanation:
the general equation of a circle is:
# x^2 + y^2 + 2gx + 2fy + c = 0# To get the equation given here into this form , require to divide through by 2.
( dividing by 2 ) :
# x^2 + y^2 + 2x - 4y - 11 = 0# Comparing this equation to the general one we can extract
2g = 2 → g = 1 , 2f = -4 → f = -2 and c = - 11
from this we obtain : centre = (-g , -f ) = (-1 , 2 )
and radius
# r = sqrt(g^2+f^2 - c ) = sqrt(1^2+(-2)^2 + 11) = 4# equation of a circle in standard form is :
# (x-a)^2 +(y-b)^2 = r^2 #
where (a , b ) are the coords of centre and r is the radiusstandard form :
# (x+1)^2 + (y-2)^2 = 16#