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    a circle passes through A(3,1) b(8,2) C(2,6)

    A) Find the point of intersection of the perpendicular bisectors of AB and BC

    do I just find the mid points and gradients of AB and BC and the take the negative reciprocal of the gradient and form an equation for AB and AC using y-1=m(x-x1) and solve them simultaneously to get the points of intersection? thanks
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    yes
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    (Original post by L'Evil Fish)
    yes
    thanks
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    well since these are points on a circle the bisectors will be diameters, so it is asking you to find the center of the circle ?
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    (Original post by L'Evil Fish)
    yes
    hey fish how do I do this?

    a circle touches the positive x-axis and y axes and its centre lies on the line y=3x-4

    find an equation of the circle
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    (Original post by madfish)
    hey fish how do I do this?

    a circle touches the positive x-axis and y axes and its centre lies on the line y=3x-4

    find an equation of the circle
    What is the general equation of a circle?
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    (Original post by Noble.)
    What is the general equation/form of a circle?
    (x-a)^2+(y-b)^2=r^2
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    (Original post by madfish)
    (x-a)^2+(y-b)^2=r^2
    Ok, what do you know about the point (a,b). Also, how can you use the fact it touches both axes?
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    (Original post by Noble.)
    Ok, what do you know about the point (a,b). Also, how can you use the fact it touches both axes?
    (a,b) lies on the line y=3x -4
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    (Original post by madfish)
    (a,b) lies on the line y=3x -4
    Yep. Also, given that it touches the axes - what does this mean?
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    (Original post by Noble.)
    Yep. Also, given that it touches the axes - what does this mean?
    um, I am not too sure. :/
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    (Original post by Noble.)
    Yep. Also, given that it touches the axes - what does this mean?
    something about the radius?
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    (Original post by madfish)
    um, I am not too sure. :/
    Well, if it touches both axes we know that there is a solution when x=0 and when y=0.

    This is much more easily visualised if you draw the line y=3x-4 and you essentially are looking for a point on the line where you can draw a circle and have it just touch both axes - this point will be unique. However, for the equation y=x you should be able to see there are an infinite number of solutions. For example, if the centre was (1,1) and you had the radius as 1 it would satisfy the problem, and in fact any centre of (p,p) with radius p works - there are infinitely many solutions. However, because this line is not going through the origin we're looking for a point on the line y=3x-4 which is at equal distances from the y-axis and the x-axis. Again, if you don't understand this, you need to draw it. Clearly, the centre (5,11) is on the line y=3x-4 - but can we form a circle which touches both the x-axis and y-axis?
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    (Original post by Noble.)
    Well, if it touches both axes we know that there is a solution when x=0 and when y=0.

    This is much more easily visualised if you draw the line y=3x-4 and you essentially are looking for a point on the line where you can draw a circle and have it just touch both axes - this point will be unique. However, for the equation y=x you should be able to see there are an infinite number of solutions. For example, if the centre was (1,1) and you had the radius as 1 it would satisfy the problem, and in fact any centre of (p,p) with radius p works - there are infinitely many solutions. However, because this line is not going through the origin we're looking for a point on the line y=3x-4 which is at equal distances from the y-axis and the x-axis. Again, if you don't understand this, you need to draw it. Clearly, the centre (5,11) is on the line y=3x-4 - but can we form a circle which touches both the x-axis and y-axis?
    okay I am so lost

    how did you deduce the centre?
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    (Original post by madfish)
    okay I am so lost

    how did you deduce the centre?
    What do you mean how do I deduce the centre? The equation of a circle:

    (x-a)^2 + (x-b)^2 = r^2

    has centre (a,b) and radius r

    We know the centre lies on the line y=3x-4 which implies that b = 3a-4
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    (Original post by Noble.)
    What do you mean how do I deduce the centre? The equation of a circle:

    (x-a)^2 + (x-b)^2 = r^2

    has centre (a,b) and radius r

    We know the centre lies on the line y=3x-4 which implies that b = 3a-4
    okay where do we go from here?:
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    (Original post by madfish)
    okay I am so lost

    how did you deduce the centre?
    He didn't. He simply gave an example of a point that lies on the line y = 3x -4.
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    (Original post by madfish)
    okay where do we go from here?:
    Honestly, there's no point in me spoon feeding you all the way to the answer, it doesn't help you.

    You know that the circle 'touches' the x-axis and the y-axis - do you know what this means?
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    (Original post by joostan)
    He didn't. He simply gave an example of a point that lies on the line y = 3x -4.
    ahh i see
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    (Original post by joostan)
    He didn't. He simply gave an example of a point that lies on the line y = 3x -4.
    so do you set y to 0 in the line equation to find the point the circle touches the x axis?
 
 
 
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