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$\Leftrightarrow (x^2+cx) +(y^2-dy) +b =0$\\  $\Leftrightarrow (x+\frac{1}{2}c)^2 + (y-\frac{1}{2}d)^2 = -b + \frac{1}{4}(c^2+d^2)$ by completing the square\\  The equation above is a circle with a radius of $\sqrt{-b + \frac{1}{4}(c^2+d^2)}.$\\  In order for a solution to exist r^2 must be greater or equal to 0.