x^2 + y^2 + 4x - 6y + 3 = 0 and the line 2x - 4y + 1 = 0 and touching the line
x + 3y - 2 = 0 .
更新1:
x^2 + y^2 + 4x - 6y + 3 = 0 (x + 2)^2 + (y - 3)^2 = 10 centre = (-2, 3) radius = sqrt(10) 我試過計一點同一條直線既距離， absolute value of [2(-2) - 4(3) + 1]/sqrt(2^2 + 4^2) = sqrt(225/20) = sqrt(11.25) bigger than sqrt(10)
更新2:
而且，我試過用DISCRIMINANT， 2x - 4y + 1 = 0 y = (2x + 1)/4 x^2 + y^2 + 4x - 6y + 3 = 0 x^2 + (2x + 1)^2/16 + 4x - [6(2x + 1)]/4 + 3 = 0 16x^2 + (2x + 1)^2 + 64x - 48x - 24 + 48 = 0 20x^2 + 20x + 25 = 0 4x^2 + 4x + 5 =0 DISCRIMINANT = (-4)^2 - 4(5)(4) = 16 - 80 = - 64 smaller than 0
更新3:
咁如果用family of circles條式去計， 即係 x^2 + y^2 + 4x - 6y + 3 + k(2x - 4y + 1) = 0， 係咪k係imaginary number? 定係根本冇real solutions? 咁如果k係imaginary既話又係咩意思?
