Find the center and radius. X^2+y^2-2x+4y-4=0

Please help show me how to work this

The standard form of the equation of a circle is:
(x-h)² + (y-k)² = r², where (h,k) is the center and r is the radius.

So, write the equation in standard form by completing the square:
x² + y² - 2x + 4y - 4 = 0
x² + y² - 2x + 4y = 4
x² - 2x + y² + 4y = 4
x² - 2x + (-2/2)² + y² + 4y + (4/2)² = 4 + (-2/2)² + (4/2)²
x² - 2x + (-1)² + y² + 4y + 2² = 4 + (-1)² + 2²
x² - 2x + -1² + y² + 4y + 4 = 4 + 1 + 4
(x-1)² + (y+2)² = 9
(x-1)² + (y+2)² = 3²

That equation is of the form (x-h)²+(y-k)²=r², where (h,k) is (1,-2) and r is 3.

So, the center is (1,-2) and the radius is 3.

Hope that helps! :)

separate into two groups (x and y):

(x^2 - 2x) + (y^2 + 4y) - 4 = 0

complete the squares and remember to balance the extra stuff you add in:

(x^2 - 2x + 1) + (y^2 + 4y + 4) - 4 - 1 - 4 = 0
(x - 1)^2 + (y + 2)^2 = 9

now you have it in the good form so your center is at (1, -2) and your radius is sqrt(9) = 3