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