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
From x ^ 2 + y ^ 2 + y ^ 2 + DX + y ^ 2 + DX + ey + F = 0, we can get (x +, #½; d) &\### (178, + (y + e \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\(4f + 4F) + F = 0 (x + ½ d) ² + (y