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Math Help - Help with a few circles......

  1. #1
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    Help with a few circles......

    Hey guys new to the forum hoping you can fix my stupidity. SO the first problem asks to find the equation of the circle whose center is on the y axis and that contain the points (1,4) and (-3,2). Give the center and the radius. I can find the radius and write the equation I am just struggling to find the midpoint. The other problem is similar. Find the midpoint of a circle passing through the points (3,1) (0,0) and (8,4). Any help would be greatly appreciated!
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  2. #2
    Senior Member jakncoke's Avatar
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    Re: Help with a few circles......

    So if your center is on the y axis it means the x coordinate is 0 and since the distance from center of circle to the points it contains is equal, if (0, y) was our center.

    Distance between (0,y) and (1,4) = Distance between (0,y) and (-3,2)

    So using the distance formula. Distance between (0,y) and (1,4) =  \sqrt{1 + (y-4)^2}
    Distane between (0,y) and (-3,2) is  \sqrt{9 + (y-2)^2}

    so  \sqrt{1 + (y-4)^2} =  \sqrt{9 + (y-2)^2} or  1 + (y-4)^2 =  + (y-2)^2
    which gives the identity,  -4 + 4y = 0 or y = 1

    so (0,1) is your center and  \sqrt{9 + (1-2)^2} = \sqrt(10) is your radius.
    Last edited by jakncoke; January 27th 2013 at 11:48 PM.
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  3. #3
    Senior Member jakncoke's Avatar
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    Re: Help with a few circles......

    For the second one.

    Do the same thing so you should get 2 linear equations in 2 unknown. If (x,y) were our midpoint
    Distance between (x,y) and (3,1) = Distance between (x,y) and (0,0)
    and
    Distance between (x,y) and (8,4) = Distance between (x,y) and (0,0) (Note: You could also use (3,1) here but (0,0) is easier)

    So Distance b/w (x,y) and (3,1) =  \sqrt{(x-3)^2 + (y-1)^2}
    Distance b/w (x,y) and (0,0) =  \sqrt(x^2 + y^2)
    so  \sqrt{(x-3)^2 + (y-1)^2} = \sqrt(x^2 + y^2) or  x-3)^2 + (y-1)^2 = x^2 + y^2 or  y = 5-3x

    Now doing the same for Distance b/w (x,y) and (8,4) =  \sqrt{(x-8)^2 + (y-4)^2}
    Distance b/w (x,y) and (0,0) =  \sqrt{x^2 + y^2}
    setting them equal  (x-8)^2 + (y-4)^2 = x^2 + y^2 or  y = 10 - 2x

    solve  y = 10 - 2x and  y = 5 - 3x or 10-2x = 5 - 3x and so x = -5 and y = 20
    our circle center (-5,20)
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    Re: Help with a few circles......

    I was wondering if you could check me on this next problem. Same layout. Find the circle with the points (1,-4)(-3,4)(4,5) I came up with Center (1,1) radius of 5. I would show you all my work but it I am new to the site and it would take me forever. It seems to check out because the points are all 5 units from center. Just wanted to double check with someone. Thanks in advance.
    Last edited by bearjr32; January 28th 2013 at 08:07 PM.
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  5. #5
    Senior Member jakncoke's Avatar
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    Re: Help with a few circles......

    Looks good
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