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

1 Answer
Feb 8, 2016

Answer:

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

Explanation:

The general equation of a circle is
# x^2 + y^2 + 2gx + 2fy + c = 0#

the equation here # x^2 + y^2 - 2x + 4y -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((-1)^2+2^2 + 4) = sqrt9 = 3#