In the general form #2x^2 + 2y^2 + 4x -8y -22 = 0#, how do you convert to standard form?

1 Answer
Feb 6, 2016

Answer:

# (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 radius

standard form : # (x+1)^2 + (y-2)^2 = 16#