標題:
F5 Maths circle
發問:
A straight line cuts the circle C1: (x-4)^2+(y-3)^2=64 at P (-4, 3) and Q(4, -5). Find the equation of circle C2 passing through P, Q and the centre of C1. The answer is x^2+y^2+2y-31=0 Show me the steps please. 更新: Why D = 0 ???
最佳解答:
Centre of C1: (4,3). We can set C2 x^2 + y^2 + Dx + Ey + F = 0 Sub. P (-4, 3) and Q(4, -5) and M(4, 3) 25 - 4D + 3E + F = 0...(1) 41 + 4D - 5E + F = 0...(2) 25 + 4D + 3E + F = 0...(3) (1) (3)=> D = 0 Then 25 + 3E + F = 0...(1) 41 - 5E + F = 0...(2) => E = 2, F = -31 So, C2: x^2 + y^2 + 2y - 31 = 0 2011-01-27 17:26:02 補充： (3) - (1) 8D = 0 D = 0
其他解答:
F5 Maths circle
發問:
A straight line cuts the circle C1: (x-4)^2+(y-3)^2=64 at P (-4, 3) and Q(4, -5). Find the equation of circle C2 passing through P, Q and the centre of C1. The answer is x^2+y^2+2y-31=0 Show me the steps please. 更新: Why D = 0 ???
最佳解答:
Centre of C1: (4,3). We can set C2 x^2 + y^2 + Dx + Ey + F = 0 Sub. P (-4, 3) and Q(4, -5) and M(4, 3) 25 - 4D + 3E + F = 0...(1) 41 + 4D - 5E + F = 0...(2) 25 + 4D + 3E + F = 0...(3) (1) (3)=> D = 0 Then 25 + 3E + F = 0...(1) 41 - 5E + F = 0...(2) => E = 2, F = -31 So, C2: x^2 + y^2 + 2y - 31 = 0 2011-01-27 17:26:02 補充： (3) - (1) 8D = 0 D = 0
其他解答:
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