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匿名 發問於 科學及數學數學 · 6 年前

Geometry question 2

How to do the following ?

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更新:

Please go to the site below to solve Geometry question 4 if you can:

https://hk.knowledge.yahoo.com/question/question?q...

2 個解答

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  • 土扁
    Lv 7
    6 年前
    最愛解答

    3.

    Put y = 0 (x-axis) into the equation (x + 3)² +(y - 4)² = 25

    (x + 3)² + (0 - 4)² =25

    (x + 3)² + 16 = 25

    (x + 3)² = 9

    x + 3 = 3 or x + 3 = -3

    x = 0 and x = -6

    Hence, OA = 0 - (-6) = 6

    Put x = 0 (y-axis) into the equation (x + 3)² +(y - 4)² = 25

    (0 + 3)² + (y - 4)² =25

    9 + (y - 4)² = 25

    (y - 4)² = 16

    y - 4 = 4 or y - 4 = -4

    y = 8 or y = 0

    Hence, OC = 8 - 0 = 8

    Area of rectangle OABC

    = OA * OC

    = 6 * 8

    = 48 ...... answer : B

    ====

    4.

    Let O(a, b) be the centre of the circle.

    The circle touches the positive x-axis at C(a, 0).

    Radius of the circle = b

    OA = b :

    √[(a - 0)² + (b - 8)²] =b

    (a)² + (b - 8)² = b²

    a² + b² -16b + 64 = b²

    a² - 16b + 64 = 0 ...... [1]

    OB = b :

    √[(a - 0)² + (b - 2)²] =b

    (a)² + (b - 2)² = b²

    a² + b² - 4b+ 4 = b²

    a² - 4b + 4 = 0 ...... [2]

    [1] - [2] :

    -12b + 60 = 0

    b = 5

    Put into [1] :

    a² - 16(5) + 64 = 0

    a² = 16

    a = 4 or a = -4 (rejected)

    Centre = (4, 5) and radius = 5

    The equation :

    (x - 4)² + (y - 5)² =5²

    x² - 8x + 16 + y² -10y + 25 = 25

    x² + y² -8x - 10y + 16 = 0 ...... answer : (A)

    2015-03-04 21:59:46 補充:

    4. Alternative answer :

    Let x² + y² + Dx + Ey + F = 0 be the required equation.

    Centre (-D/2, -E/2) is in the first quadrant.

    -D/2 > 0 and -E/2 > 0

    Hence, D < 0 and E < 0 ...... [1]

    The circle touches x-axis but not y-axis.

    Hence, D not equatl to E ...... [2]

    The answer : A

  • 6 年前

    3.

    Put x = 0,

    (y - 4)² = 25 - 9 = 16

    y - 4 = 4 or -4

    y = 0 or 8

    OC = 8

    Put y = 0,

    (x + 3)² + 16 = 25

    (x + 3)² = 9

    x + 3 = 3 or -3

    x = 0 or -6

    OA = 6

    Area = 8 * 6 = 48

    (B)

    2015-03-04 21:51:09 補充:

    4.

    Let the centre be C = (a, b)

    By symmetry,

    b = (2 + 8)/2 = 5

    Then, we note that radius = 5 also.

    This is because the circle touches x-axis.

    Denote the mid-point of AB be M.

    Consider right-angled triangle AMC.

    3² + a² = (radius)²

    9 + a² = 25

    a = 4

    2015-03-04 21:51:28 補充:

    With centre (4, 5) and radius 5, the equation is

    (x - 4)² + (y - 5)² = 5²

    x² + y² - 8x - 10y + 16 = 0

    (A)

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