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    (Original post by TenOfThem)
    The 2a is only the radius not the distance to the origin
    Ah yes, my tiredness I think. I thought origin = radius Thanks
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    yes, as I said the radius is 2a which is what you got
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    (Original post by MedicalMayhem)
    Ah yes, my tiredness I think. I thought origin = radius Thanks
    That was why I wanted you to picture it

    If you use Pythag to find the centre to the origin and use radius = 2a then you will have the distance from the circle to the origin


    If you picture it you will have a lie that goes from the centre to the origin


    For the distance to the y-axis do the same thing ... how far is the centre from the axis so haw far is the circle from the axis
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    (Original post by TenOfThem)
    yes, as I said the radius is 2a which is what you got
    Hm.. I still didn't get 3a :/ Well through pythagoras (and the origin being 0,0) I did

    (4a-0)^2 + (3a-0)^2 = 25a^2

    So root 25a^2 = 5a?

    Have I just missed something out again? :/
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    No that is right

    So

    centre to circle = 2a (radius)
    centre to origin = 5a

    circle to origin =
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    (Original post by TenOfThem)
    That was why I wanted you to picture it

    If you use Pythag to find the centre to the origin and use radius = 2a then you will have the distance from the circle to the origin


    If you picture it you will have a lie that goes from the centre to the origin


    For the distance to the y-axis do the same thing ... how far is the centre from the axis so haw far is the circle from the axis
    :/ I don't see how the radius is related to the distance from centre to the origin? I understand the line, but it won't be in proportion as we have no real values?

    But wouldn't visualising it only give you an estimate? :/
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    (Original post by TenOfThem)
    No that is right

    So

    centre to circle = 2a (radius)
    centre to origin = 5a

    circle to origin =
    Oh yeah! So that's why you said to picture it!
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    Picture the line from the centre to the origin ... consider where it crosses the circle
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    (Original post by MedicalMayhem)
    Oh yeah! So that's why you said to picture it!
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    I am off to bed now MM

    Night

    You are doing well with the algebra elements which is where a lot would fall down

    More practice on what to look for and these will be a doddle for you
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    (Original post by TenOfThem)
    I am off to bed now MM

    Night

    You are doing well with the algebra elements which is where a lot would fall down

    More practice on what to look for and these will be a doddle for you
    Yeah I'm off too, I just attempted c) and I can't see how you can find out anything about the tangent from the information given. As you have the centre of the circle which is (4a,-3a).

    Radius is 2a. Point where they intercept is at 4a,-a.

    The only way I think you can do it, which also makes sense, is that as the point of intercept is directly below the center of the circle, then and is perpendicular to the normal (which is infinitely steep) it would have a gradient of 0?

    So you do 0 (x-4a) = y+a. So the tangent is y=-a?

    Thanks so much for all your help tonight, and clearly you can tell that you are a great teacher, and you have one quality which many often lack :rolleyes: ...Patience, and I really am appreciative of your perseverence in helping me understand, and I think that I may not be a lost cause in maths (as I began to believe after loads of unanswerable questions from the textbook!)
 
 
 
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