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

1 Answer
May 30, 2016

Answer:

Put the equation into standard form to find it to be a circle with radius #3# and center #(-1, 2)#

Explanation:

In order to put the equation of a conic section into standard form (at which point we can see most of its properties), we use a process called completing the square on both variables.

#x^2+y^2+2x-4y=4#

#=>(x^2+2x+1) - 1 + (y^2 - 4y + 4) - 4 = 4#

#=> (x+1)^2 + (y-2)^2 = 4+4+1 = 9#

The standard form of the equation of a circle with radius #r# centered at #(x_0, y_0)# is

#(x-x_0)^2 + (y-y_0)^2 = r^2#

Thus, as our equation may be written as

#(x-(-1))^2+(y-2)^2=3^2#

it represents a circle of radius #3#, centered at #(-1,2)#