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    A) a circle has its centre at the origin and cuts the x-axis at the points A(-c,0) and B(c,0). Write down the equation of the circle.

    B) The point P (h,k) lies on the circle. Find expressions for the gradients of AP and BP.

    C) By showing that the product of the gradients of AP and BP is -1, prove that <APB= 90

    A) x^2 + y^2 =C^2

    B) Gradient of AP = K/H+C
    Gradient BP = K/H-C

    C) this is the part im stuck on, I don't know if this is the correct answer I times the gradient of AP and BP to get K^2/H^2-C^2 = -1

    By doing that have I proved <APB = 90?
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    (Original post by Vorsah)
    A) a circle has its centre at the origin and cuts the x-axis at the points A(-c,0) and B(c,0). Write down the equation of the circle.

    B) The point P (h,k) lies on the circle. Find expressions for the gradients of AP and BP.

    C) By showing that the product of the gradients of AP and BP is -1, prove that <APB= 90

    A) x^2 + y^2 =C^2

    B) Gradient of AP = K/H+C
    Gradient BP = K/H-C

    C) this is the part im stuck on, I don't know if this is the correct answer I times the gradient of AP and BP to get K^2/H^2-C^2 = -1

    By doing that have I proved <APB = 90?
    Well if two lines AP and BP are perpendicular what does it say about angle APB?!
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    (Original post by Mr M)
    Well if two lines AP and BP are perpendicular what does it say about angle APB?!
    It's 90
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    (Original post by Vorsah)
    It's 90
    And do you understand that lines are perpendicular if the product of their gradients is -1?
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    (Original post by Mr M)
    And do you understand that lines are perpendicular if the product of their gradients is -1?
    Yes
    • Community Assistant
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    (Original post by Vorsah)
    Yes
    So you are happy now?
 
 
 
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