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    Just a few, probably the first steps that I cant do....


    1) Find the equation for the circle that passes through the points (2,0), (8,0) and (5,9).

    2) Find in cartesian form, an equation for the circle which touches the y axis at the point (0,6) and also passes through the point (1,3)

    3) 4x^2 + 4y^2 - 20x - 73 = 0 - find the centre and radius

    4) x = sec t, y = tan t - Find an eqation in cartesian form for these pararmetric equations.

    For number for I seaid:

    x = 1/tan t and so x = 1/y

    But it says that ... x^2 - y^2 = 1???
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    (Original post by ianripping)
    Just a few, probably the first steps that I cant do....
    1) Find the equation for the circle that passes through the points (2,0), (8,0) and (5,9).
    equation of a circle is in the form:

    x^2 + y^2 + 2fx + 2gy + c = 0

    where the centre is (-f,-g), and r^2= (f^2 + g^2 - c)

    just substitute each set of coordiantes to get 3 equations, and solve simultaneously
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    (Original post by ianripping)

    2) Find in cartesian form, an equation for the circle which touches the y axis at the point (0,6) and also passes through the point (1,3)
    see my solution here: http://www.uk-learning.net/showthrea...975#post747975
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    for q1
    just put the values of all x and y coordinates in the equations
    x² + y ² + 2gx + 2fx + c = 0
    get three equations and solve

    for q2
    the radius is 6,as stated by the question since it touches the y axis at (0,6)

    for q3
    divide all with 4 first
    then use r = root (g² + f² - c)

    for q4
    the coeeficient for both parametric is 1
    therefore the radius is 1
    and since it doesnt get added or deducted or added with anything
    the centre is (0,0)
    thus
    x² + y² = 1
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    (Original post by ianripping)

    3) 4x^2 + 4y^2 - 20x - 73 = 0 - find the centre and radius
    divide by 4:
    x^2 + y^2 - 5x - (73/4) = 0

    now either complete the sqaure, or use the equation of circle form i gave in question 1)
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    1) Find the equation for the circle that passes through the points (2,0), (8,0) and (5,9).

    do you get centre as (5, 7.7) and radius 7.8

    or have i don somfin wrong? ....for 1
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    No I have done q1 now and got that the radius is 5, centre is 5,4
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    4) tan^2(t) + 1 = sec^2(t). Therefore, if x = sec t, y = tan t we have y^2 + 1 = x^2 which is x^2 - y^2 = 1.
 
 
 

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