Excuse me, the general equation of this circle x + y + DX + ey + F = 0, why D + e-4f > 0? How is d + e-4f > 0 obtained?

If the general equation of circle is x + y + DX + ey + F = 0, then the standard equation of circle is: (X-D / 2) ^ 2 + (y-e / 2) ^ 2 = D ^ 2 / 4 + e ^ 2 / 4-f. then according to the radius r > 0, we can know: D ^ 2 / 4 + e ^ 2 / 4-f > 0

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1. In the equation x ^ 2 + y ^ 2 + DX + ey + F = 0, if d ^ 2 = e ^ 2 = 4f, then the position of the circle () A. The length of the chord obtained by cutting two axes is equal B. Tangent to both axes C. Away from two axes D. All of the above are possible
2. If the center of x ^ 2 + y ^ 2 + DX + ey + F = 0 (d ^ 2 + e ^ 2-4f > 0) is on the straight line x + y = 0, then the relation of D, e, f What does F have to do with them
3. What are the conditions for the intersection of the circle x ＾ 2 + y ＾ 2 + DX + ey + F = 0 and the Y axis on both sides of the origin Detailed process, thank you
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5. The circle X & sup2; + Y & sup2; + DX + ey + F = 0 is symmetric with respect to the line y = X A.D+E=0 B.D-E=0.C.D+F=0 D.D-F=O
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11. What is the condition that "d square + e square - 4f > 0" is the equation x square + y square + DX + ey + F representing a circle?
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