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標題:

F.5 MATHS(Locas) 15 分

發問:

Find the equation of a circle passing through the origin and with centre at the point of intersection of the lines 2x +y =4 and y- x +5=0 ge solution

最佳解答:

2x +y = 4........(1) y - x +5 = 0.....(2) (1) - (2), 2x + y - y + x - 5 = 4 3x = 9 x = 3 So y = x - 5 = 3 - 5 = -2 The coordinates of the centre are (3, -2) The circle passes through the origin. Thus radius = √[(3 - 0)^2 + (-2 - 0)^2] = √[9 + 4] = √13 The equation of the circle is (x - 3)^2 + (y + 2)^2 = [√13]^2 x^2 - 6x + 9 + y^2 + 4y + 4 = 13 x^2 + y^2 - 6x + 4y = 0

其他解答:

2x +y = 4........(1) y - x +5 = 0.....(2) (1) - (2), 2x + y - y + x - 5 = 4 3x = 9 x = 3 So y = x - 5 = 3 - 5 = -2 The coordinates of the centre are (3, -2) The circle passes through the origin. Thus radius = √[(3 - 0)^2 + (-2 - 0)^2] = √[9 + 4] = √13 The equation of the circle is (x - 3)^2 + (y + 2)^2 = [√13]^2 x^2 - 6x + 9 + y^2 + 4y + 4 = 13 x^2 + y^2 - 6x + 4y = 0|||||[1] 2x+y=4 [2] y-x=0 Fm [2], y=x[3] sub[3]in [1], 2y+y=4 3y=4 y=4/3 ~~ x=4/3 The centre of circle (4/3, 4/3) r = [(0-4/3)^2+ (0-4/3)^2]^1/2 r = (32/9)^1/2 r = [32^1/2] / 3 The eq. of the circle: x^2 +y^2 +2(4/3) +2(4/3)+4/3^2+4/3^2-{[32^1/2]/3}^2=0 x^2+y^2+16/3-64/3=0 x^2+y^2-58/3=06CC7293C79127CE5