# How do you convert the general form of the equation of a circle 2x^2+2y^2+4y=0 to standard form?

Jan 18, 2016

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

#### Explanation:

The standard form is : ${\left(x - a\right)}^{2} + {\left(y - b\right)}^{2} = {r}^{2}$

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

the one here : $2 {x}^{2} + 2 {y}^{2} + 4 y = 0$

(divide both sides by 2 ) : ${x}^{2} + {y}^{2} + 2 y = 0$

( comparing this to the general form ) : g = 0 , 2f = 2 so f = 1
and c = 0 .

centre = ( - g , - f ) = ( 0 ,- 1 ) and

$r = \sqrt{{g}^{2} + {f}^{2} - c} = \sqrt{0 + {1}^{2} - 0} = \sqrt{1} = 1$

equation in standard form is ; ${\left(x - 0\right)}^{2} + {\left(y + 1\right)}^{2} = {1}^{2}$

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