設一圓通過兩點A(5,1)，B(3,1)且圓心在直線x+2y-3=0，求此圓方程式： (x−(1))平方+(y+(3)(2))平'方=(5)(6)/(4) (1)(2)(3)(4)(5)(6)分別=?
回答 (1)
2017-04-02 6:35 pm
✔ 最佳答案
A,B中垂線: x = 4 與 x + 2y = 3 交點為 O = (4- 1/2) = 圓心R^2 = AO^2 = BO^2 = 1 + 9/4 = 13/4
==>> (x - 4)^2 + (y + 1/2)^2 = 13/4
(1) = 4, (3) = -1, (2) = /2, (5) = 1, (6) = 3, (4) = 4
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