Rewrite the equation of the circle (x-0.8)^2+(y+0.6)^2=0.2 in general form

x^2+y^2-1.6x+1.2y+0.8=0



Work Shown:

Expand out the x expression first
(x-0.8)^2 = (x-0.8)(x-0.8)
(x-0.8)^2 = x(x-0.8)-0.8(x-0.8)
(x-0.8)^2 = x^2-0.8x-0.8x+0.64
(x-0.8)^2 = x^2-1.6x+0.64

Then do the y 
(y+0.6)^2 = (y+0.6)(y+0.6)
(y+0.6)^2 = y(y+0.6)+0.6(y+0.6)
(y+0.6)^2 = y^2+0.6y+0.6y+0.36
(y+0.6)^2 = y^2+1.2y+0.36

So overall we can say the following
(x-0.8)^2+(y+0.6)^2=0.2 
x^2-1.6x+0.64+(y+0.6)^2=0.2 
x^2-1.6x+0.64+y^2+1.2y+0.36=0.2 
x^2+y^2-1.6x+1.2y+0.64+0.36=0.2 
x^2+y^2-1.6x+1.2y+0.64+0.36-0.2=0
x^2+y^2-1.6x+1.2y+0.8=0
which is the final answer