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cartesian equation circle question Watch

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    can someone help me solve this??
    :show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius
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    (Original post by Carokelly123)
    can someone help me solve this??
    :show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius
    Is that the whole question
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    (Original post by Carokelly123)
    can someone help me solve this??
    :show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius
    What circle?
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    (Original post by 123Master321)
    Is that the whole question
    yeah it is!
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    (Original post by Carokelly123)
    yeah it is!
    Not enough information to define a circle.
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    (Original post by RDKGames)
    Not enough information to define a circle.
    Let me math a sec
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    (Original post by Carokelly123)
    can someone help me solve this??
    :show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius
    Note that  y/x=t . Sub that expression for t into one of the expressions for x or y and then rearrange and you get the equation for a circle.
    It's just converting from parametric to Cartesian.
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    (Original post by Carokelly123)
    can someone help me solve this??
    :show that the point (1/(1+t^2) , t/(1+t^2)) lies on a circle and hence find its centre and radius
    (\frac{1}{1+t^2}, \frac{t}{1+t^2}) = (X, Y)

    Now you have two equations:

    X = \frac{1}{1+t^2},
    Y = \frac{t}{1+t^2}

    Rearrange one or both of the equations to get a function for t in terms of either X. Y or both, then substitute that function for t in the other. If the resulting equation is of the form:

    (Y-a)^2+(X-b)^2 = r^2

    Then you have a circle, centre (a,b) and radius r

    EDIT: Note there is an easy method for this particular example to do the simultaneous equations.
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    t=y/x, substitute in x=1/(1+t^2) and rearrange to get x^2 - x +y^2 = 0, then complete the square to get (x-1/2)^2 + y^2 = (1/2)^2, i.e. a circle centered on (1/2, 0) with a radius of 1/2.
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    (Original post by RogerOxon)
    t=y/x, substitute in x=1/(1+t^2) and rearrange to get x^2 - x +y^2 = 0, then complete the square to get (x-1/2)^2 + y^2 = (1/2)^2, i.e. a circle centered on (1/2, 0) with a radius of 1/2.
    Please, no full solutions.
 
 
 
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