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

Feb 6, 2016

${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 16$

Explanation:

the general equation of a circle is:

${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

To get the equation given here into this form , require to divide through by 2.

( dividing by 2 ) : ${x}^{2} + {y}^{2} + 2 x - 4 y - 11 = 0$

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 $r = \sqrt{{g}^{2} + {f}^{2} - c} = \sqrt{{1}^{2} + {\left(- 2\right)}^{2} + 11} = 4$

equation of a circle in standard form is :

${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$
where (a , b ) are the coords of centre and r is the radius

standard form : ${\left(x + 1\right)}^{2} + {\left(y - 2\right)}^{2} = 16$