(x+2)^2+(y-2)^2= 4 WRITE IN GENERAL FORM

Question

My points are (-2,4) (0,2) (-2,0) and (-4,2)

(x-h)^2+(y-k)^2=r^2

I did the end points (-2,4) and (-2,0)

r = Radical - (-2-(-2))^2+(0-4)^2/ 2 = 2

I got (x+2)^2+(y-2)^2= 4

I now have to put it in GENERAL FORM but I am so confused. I ended up with x^2+y^2+4x-4y+1=0

but I do not know why I got that answer incorrect.  PLEASE HELP?!!?

Answer 1

Given Equation of The Circle 

(x + 2)^2 + (y - 2)^2  =  4

Expand them

(x^2 + 4x + 4) + (y^2 - 4y + 4)  =  4

x^2 + 4x + 4 + y^2 - 4y + 4 - 4  =  0

x^2 + y^2 + 4x - 4y + 4  =  0

Answer :

Genral form is x^2 + y^2 + 4x - 4y + 4  =  0.