YTC 發問於 科學及數學數學 · 5 年 前

F.5 Maths equation of circle

Show that the 2 circles C1: x^2 +y^2 -16x-12y+75=0 and

C2: x^2 +y^2 +2x+12y-63=0 touch each other.

My calculation:

http://postimg.org/image/6rxfgpbfx/

I have some questions about my explanation(the red rectangle part).

I think that it is not detail enough. Also, the explanation only suitable for the case

that the circles touh externally. When they touch internally, the explanation is

wrong. Therefore, I want to ask for a better solution for this question, thank you!!

1 個解答

評分
  • 土扁
    Lv 7
    5 年 前
    最佳解答

    Let O1 and O2 be the centres of the C1 and C2 respectively.

    C1 : x² + y² - 16x - 12y + 75 = 0

    Coordinates of O1 = (16/2, 12/2) = (8, 6)

    Radius of C1 = √[(-16/2)² + (-12/2)² - 75] = 5

    C2 : x² + y² + 2x + 12y - 63= 0

    Coordinates of O 2 = (-2/2, -12/2) = (-1, -6)

    Radius of C2 = √[(2/2)² + (12/2)² - (-63)]= 10

    O1O2 = √[(8 + 1)² +(6 + 6)²] = 15

    Radius of C1 + Radius of C2 = 15

    The distance between the two centres isequal to the sum of the two radii.

    Hence, the two circles touchexternally.

    2015-03-18 20:06:13 補充:

    Alternative method :

    C1 : x² + y² - 16x - 12y + 75 = 0 ...... [1]

    C2 : x² + y² + 2x + 12y - 63 = 0 ...... [2]

    [2] - [1] :

    18x + 24y - 138 = 0

    3x + 4y - 23 = 0

    x = (23 - 4y)/3 ...... [3]

    2015-03-18 20:06:40 補充:

    Put [3] into [2] :

    [(23 - 4y)/3]² + y² + 2[(23 - 4y)/3] + 12y - 63 = 0

    (23 - 4y)² + 9y² + 6(23 - 4y) + 108y - 567 = 0

    529 - 184y + 16y² + 9y² + 138 - 24y + 108y - 567 = 0

    25y² - 100y + 100 = 0

    y² - 4y + 4 = 0

    (y - 2)² = 0

    y = 2 (double root)

    2015-03-18 20:06:59 補充:

    The two circles meet at only one point.

    Hence, the two circles touch each other.

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