The range of values of k such that the equation C: x^2 + y^2 + 2x +6y - 2k = 0 represents a real circle or a point circle.
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2007-11-04 8:21 pm
回答 (2)
2007-11-04 8:32 pm
✔ 最佳答案
Radius of C = sqrt[(-1)^2 + (-3)^2 - (-2k)] >= 01+9+2k >= 0
k >= -5
For the circle, x^2 + y^2 - 2x + 4y + 8 = 0
Radius = sqrt[(1)^2 + (-2)^2 - 8]
=sqrt(1+4-8)
=sqrt(-3), which is undefined.
Hence the circle is imaginary.
2007-11-04 8:43 pm
√(-2/2)^2 + (-6/2)^2 - (-2k)= Real Number
√10+2k ≧Real Number = 0
k≧-5
Show that the equation C; x^2 + y^2 - 2x + 4y + 8 = 0 represents an imaginary circle
√(-2/-2)^2 + (-4/2)^2 - (8)= √-3 = √3i，which is an imaginary circle.
√10+2k ≧Real Number = 0
k≧-5
Show that the equation C; x^2 + y^2 - 2x + 4y + 8 = 0 represents an imaginary circle
√(-2/-2)^2 + (-4/2)^2 - (8)= √-3 = √3i，which is an imaginary circle.
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