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    hi

    I'm having real trouble with these sorts of questions..please help!

    1) AB is diameter of a circle
    A = (1,0) B = (3,0)
    Find equation of the circle

    2) Find equation of the circle passing through the points:
    (0,0) (0,4) (6,0)

    Thanks x
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    (Original post by jazzyj)
    hi

    I'm having real trouble with these sorts of questions..please help!

    1) AB is diameter of a circle
    A = (1,0) B = (3,0)
    Find equation of the circle

    2) Find equation of the circle passing through the points:
    (0,0) (0,4) (6,0)

    Thanks x
    erm
    you sure thats the ony question?
    looks a bit too simple to be true
    anyways
    the radius for question 1 is obviously 1
    and then try finding the center
    its the centre of the 2 points
    thus

    its
    (x-2)² + y² = 1
    i think...

    now for q2
    take A(0,0) , B(0,4) , C(6,0)

    use the equation
    x² + y² + 2gx + 2fx + c = 0

    the equations are
    c = 0 ----(1)
    16 + 8f + c = 0 ----(2)
    36 + 12g + c = 0 -----(3)

    solve them all

    you`ll get f as -2
    g as -3

    thus...equate them al in the original equation
    x² + y² - 6x - 4x = 0


    I think thats the answers
    sorry if im wrong
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    the answer to q1 is supposed to be (x-1)(x-3) + y^2 = 0

    and for q2, (x-3)^2 + (y-2)^2 = 13
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    (Original post by jazzyj)
    the answer to q1 is supposed to be (x-1)(x-3) + y^2 = 0
    that is the same as (x-2)^2 + y^2 = 1 , try expanding and you'll see
    qusetion 2 is the same, if you factorise the answers given by MalaysianDude, you'll see that they are in the same form as the answers you want
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    ah thank you!
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    Alternative solution
    ============

    Write the points as,
    O(0,0), A(0,4), B(6,0)

    Now, if O, A and B all lie on the circle then OA and OB are chords of the circle. That means that the perpindiculars to the mid-points of the chords pass through the centre of the circle.
    The perpindicular passing thro' the mid-point of OA is the line y=2.
    The perpindicular passing thro' the mid-point of OB is the line x=3.
    Thes two lines, or perpindiculars, intersect at the point (3,2) - the centre of the circle!
    Since the centre of the circle is the point C(3,2), and the origin, O, is a point on the circle, then pythag gives the radius², OC², as 3² + 2² = 13.
    So, eqn of circle is,

    (x-3)² + (y-2)² = 13
    =============
 
 
 

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