Describing the shape of the curve

I have an equation of the the following function:
d(x^2+y^2) + 2(ax-by)+c = 0 where a,b, c and d are reals.
Could someone please explain how I could explain the shape of the curve when d=0 and d is not equal to 0.
For d=0 we get a line.
But for d not equal to 0 do we get a circle? I guess I could complete the square for x and y and so obtain an equation for a circle, is that correct?
Thank you!

But for d not equal to 0 do we get a circle? I guess I could complete the square for x and y and so obtain an equation for a circle, is that correct?

I would add a caveat to what NotNotBatman has said.

For $d\not=0$, if you rearrange into the form $(x-p)^2+(y-q)^2=r^2$, then what's on the righthand side needs to be positive for a circle. If it's zero, your equation defines a point (circle radius zero), and if it's negative, there's no real solution.