更新1:
I calculated (a). The equation of C is x^2+y^2-8x-12y+27=0
更新2:
(b) my steps y=-1/2k+C Sub. y=0 C=1/2k Therefore the equation is y=-1/2x+1/2k I did wrong or right?
Maths problem: how to do (b), thanks.?
回答 (1)
2021-03-22 8:29 am
✔ 最佳答案
圓心 (4,6) 與直線 y = 1 的距離是 |6-1| = 5.所以圓 C 方程式是
(x-4)^2 + (y-6)^2 = 5^2
L 斜率 -1/2, x-截距 k, 方程式
y = (-1/2)(x-k)
交圓 C 於 A, B, 所以
(x-4)^2+ [-1/2(x-k)-6]^2 = 25
= x^2-8x+16 +x^2/4 -2(x/2)(k/2-6)+(k/2-6)^2
∴ (5/4)x^2 - (k/2+2)x + (k/2-6)^2-9 = 0
x = {(2+k/2) ± √[125-(k-16)^2]}/(5/2)
∴ A, B 之中點座標:
x = (2+k/2)/(5/2) = (4+k)/5
y = (-1/2)[(4+k)/5-k] = 2(k-1)/5
(計算正確與否請自己檢查.)
收錄日期: 2021-05-04 00:40:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20210321154337AAENX5v