美夢成真教甄討論區

若圓C： x2 + y2 = 9, 已知一直線過點 (3,1) 與圓C相切且斜率為實數m, 則m = (A) 1 (B) 2 (C) －4/3 (D) － 5/4

令該直線為 y = m(x - 3) + 1 = mx - (3m - 1)
由於相切，x^2 + [mx - (3m - 1)]^2 = 9 之判別式為 0

x^2 + [mx - (3m - 1)]^2 = 9
(m^2 + 1)x^2 - 2m(3m - 1)x + [(3m - 1)^2 - 9] = 0
判別式 = [-2m(3m - 1)]^2 - 4(m^2 + 1)[(3m - 1)^2 - 9] = 0
......
6m + 8 = 0
m = -4/3