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    Can someone please explain to me how you do this question.

    Find the equation of the tangent to the following circle at the stated point;

    (x-1)^2+(y+2)^2=13 at point (3,1)
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    (Original post by kyra.wilson)
    Can someone please explain to me how you do this question.

    Find the equation of the tangent to the following circle at the stated point;

    (x-1)^2+(y+2)^2=13 at point (3,1)
    Have you come across implicit differentiation?
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    (Original post by SeanFM)
    Have you come across implicit differentiation?
    we've done differentiation, not sure about the implicit side !
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    Differentiate the function implicitly, you'll get dy/dx=(1-x)/(y+2), work from there.
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    (Original post by kyra.wilson)
    we've done differentiation, not sure about the implicit side !
    Okey dokey, don't worry about it then - implicit is slightly different to normal differentiation.

    It may help to sketch the circle and the tangent to that point.

    What can you say about the gradient of a line from the centre of the circle to the pont(3,1) compared to the gradient of the tangent?
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    (Original post by SeanFM)
    Okey dokey, don't worry about it then - implicit is slightly different to normal differentiation.

    It may help to sketch the circle and the tangent to that point.

    What can you say about the gradient of a line from the centre of the circle to the pont(3,1) compared to the gradient of the tangent?
    That they are perpendicular to each other!
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    (Original post by kyra.wilson)
    Can someone please explain to me how you do this question.

    Find the equation of the tangent to the following circle at the stated point;

    (x-1)^2+(y+2)^2=13 at point (3,1)
    do implicit differentiation

    basically differentiate everything but when you differentiate something in terms of y multiply it by dy/dx

    Make dy/dx the subject and then plug in the known values, from here should be obvious to even someone who is learning C1
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    (Original post by kyra.wilson)
    That they are perpendicular to each other!
    Correct, so what does that tell you about the relationship between their gradients?
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    (Original post by SeanFM)
    Correct, so what does that tell you about the relationship between their gradients?
    that its the negative reciprocal
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    (Original post by kyra.wilson)
    that its the negative reciprocal
    :borat: well done

    So do you know what to do now?
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    This is almost definitely a C1 question. No one should need to do implicit differentiation in C1...

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    (Original post by SeanFM)
    :borat: well done

    So do you know what to do now?
    to a degree but how do you find the gradient of the line from the centre to point (3,1)?
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    (Original post by kyra.wilson)
    to a degree but how do you find the gradient of the line from the centre to point (3,1)?
    You have two sets of co-ordinates and the line goes from one to the other, soo...
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    Well, you don't need implicit differentiation, we just made the mistake of assuming so.

    Alternative NON calculus method:

    Find the gradient from the centre of the circle to the point, find the negative reciprocal etc, do as you will.
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    (Original post by SeanFM)
    You have two sets of co-ordinates and the line goes from one to the other, soo...
    m=(y2-y1)/(x2-x1)????
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    (Original post by kyra.wilson)
    m=(y2-y1)/(x2-x1)????
    Correct you should have all you need now.
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    (Original post by SeanFM)
    Correct you should have all you need now.
    thank you very much, this has helped immensely
 
 
 
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