# How do you find the center and radius of x^2+y^2-2x+4y-4=0?

Feb 8, 2016

centre =(1 , -2 ) and r = 3

#### Explanation:

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

the equation here ${x}^{2} + {y}^{2} - 2 x + 4 y - 4 = 0$
is in this form and by comparison of the 2 equations , obtain

2g = - 2 → g = -1 and 2f = 4 → f = 2 , c = -4

now centre = (-g , -f ) = (1 , -2 )

and radius , r $= \sqrt{{g}^{2} + {f}^{2} - c}$

hence r$= \sqrt{{\left(- 1\right)}^{2} + {2}^{2} + 4} = \sqrt{9} = 3$