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Math Help - Finding the radius/area of a circle...

  1. #1
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    Finding the radius/area of a circle...

    Hello everyone, this is my first post. A student I used to tutor just sent me this and for some reason it has me totally stumped. I'm just not thinking about it in the right way and can't figure out the trick required to find the radius of the smaller circle.

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  2. #2
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    for want of a better word i'll call the "system" involving the small circle and the origin A. and the system involveing the origin and the large circle B.

    Are A and B similar? Just a thought.
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  3. #3
    Rhymes with Orange Chris L T521's Avatar
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    Quote Originally Posted by amroddr View Post
    Hello everyone, this is my first post. A student I used to tutor just sent me this and for some reason it has me totally stumped. I'm just not thinking about it in the right way and can't figure out the trick required to find the radius of the smaller circle.

    http://img26.imageshack.us/img26/491...xtracredit.jpg
    We don't knowingly help with assignments that count towards their final grade. See rule #6 here: Forum Rules.

    Thread closed.

    EDIT: I've decided to re-open the thread due to correspondence with the member; he failed to mention that the deadline has passed for this, and that he's just curious how to do it since it has stumped him and his fellow college algebra tutors.
    Last edited by Chris L T521; May 25th 2011 at 06:53 PM.
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  4. #4
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    Thank you for unlocking this, sorry again about not making my intentions more clear in the original post.

    I was able to solve this using trig and constrained optimization. However, I'm trying to find the solution using only things that would be in the bag of a high school algebra 2 student.

    So I worked on this a little bit more, and here's what I came up with. Going with the idea that both of these circles are tangent to both the x and y axes, and that their center points are along y=x...

    I found the point where the two circles are in contact with one another ( [√2-1]/√2 , [√2-1]/√2 )

    And, since the small circle is tangent to both axes, and has the center point along the line y=x, the values for the center point coordinates (h,k) and the value for r are the same. h=k=r

    So, I'm substituting r into the circle equation to get (x-r)^2 + (y-r)^2 = r^2.

    Then, I'm plugging in the known point coordinates ( [√2-1]/√2 , [√2-1]/√2 )

    I expected this to yield a quadratic equation with the two answers being the radius for the large circle (r=1) and then the radius for the small circle, which will then give us the area. However, I'm doing something wonky with my manipulation of the equation because I'm only getting 1 answer, not 2. The answer I am getting however, is the same as the answer I got from using trig to solve it.

    The answer I'm getting for the radius is r = (sqrt(2)-1)/(1+sqrt(2)) = 0.172

    The area is A = pi ((sqrt(2)-1)/(1+sqrt(2))^2 = 0.0925
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  5. #5
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    So I finally realized that I was being completely silly and was looking to be able to factor out h=k=r circle equation instead of just looking for the positive and negative root. Got it all worked out now. I also discovered a MUCH simpler way of doing it after reading the first reply and thinking about it. Thank you SpringFan.

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