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    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!
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    (Original post by spacewalker)
    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!
    Yes, correct.
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    (Original post by spacewalker)
    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.
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    (Original post by ghostwalker)
    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.
    Nice, I missed that.
 
 
 
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