(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1:
Why D = 0 ???
F5 Maths circle
2011-01-25 6:17 am
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.
(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.
回答 (1)
2011-01-25 6:22 am
✔ 最佳答案
Centre of C1: (4,3). We can set C2x^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
收錄日期: 2021-04-26 14:04:29
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