How do you graph x^2 + y^2 + 10x - 8y +40 = 0?

Feb 4, 2016

circle : centre = (-5,4 ) and radius = 1

Explanation:

The general equation of a circle is ${x}^{2} + {y}^{2} + 2 g x + 2 f y + c = 0$

The equation here ${x}^{2} + {y}^{2} + 10 x - 8 y + 40 = 0$

this equation 'matches' the general equation and by comparison

 2g= 10 → g=5 , 2f =- 8 → f = -4 ,c =40

centre = (-g , -f ) = (-5 , 4 )

and $r = \sqrt{{g}^{2} + {f}^{2} - c} = \sqrt{{5}^{2} + {\left(- 4\right)}^{2} = 40} = \sqrt{25 + 16 - 40} = 1$