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Thread: Solve this please

  1. #1
    Oct 2012
    long island

    Solve this please

    Solve this please-2012-10-05-23.21.01.jpg
    Find the radius of the circle at the bottom of the image!!! i do need to get it done asap so if anyone could help please
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  2. #2
    Super Member

    May 2006
    Lexington, MA (USA)

    Re: Solve this please

    Hello, midnightnurse!

    We need more information.

    Click image for larger version. 

Name:	2012-10-05 23.21.01.jpg 
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    Find the radius of the circle at the bottom of the image.

    A full description of the figure is required.
    Presently, I am guessing the following "facts" . . .

    We have a quadrant of a circle with radius 4 (the "4-circle").
    There is a semicircle of radius 2 (the "2-circle").
    The "4 cm" at the bottom is not the length of the chord.
    . . It is obviously a radius of the 4-circle.

    We have a mystery circle tangent to the 2-circle.
    It appears to extend beyond the 4-circle.

    With no further constraints, it could have any radius.
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  3. #3
    MHF Contributor
    Sep 2012
    Washington DC USA

    Re: Solve this please

    I assume the problem intends this:
    P1) Rightmost semicircle is centered on that vertical line, has diameter 4 cm.
    P2) Bottom shape is a semicircle whose center is on that horizontal line.
    P3) The circles touch at a point (hence tangentially).
    P4) The leftmost point of the bottom circle is 4 cm from the bottom-most point of the right semicircle.
    In that case, you might set the problem up this way:

    0) All distances in centimeters.
    1) Bottom left point is the origin.
    2) Bottom semicricle has radius r.
    3) Right semicircle has radius 2 and center (4, 2).

    Then by #2, bottom semicircle has center at (r, 0).
    Here's the key observation: since they touch tangentially, the distance between two centers is the sum of their radii.
    That becomes Distance( (r,0), (4,2) ) = 2 + r.

    $\displaystyle \sqrt{ (4-r)^2 + (2-0)^2} = 2+r$

    $\displaystyle \sqrt{ (4-r)^2 + 4} = 2+r$

    $\displaystyle (4-r)^2 + 4 = (2+r)^2$

    $\displaystyle (16 - 2(4)(r) + r^2) + 4 = 4 + 2(2)(r) + r^2$

    $\displaystyle r^2 - 8r + 20 = r^2 +4r + 4$

    $\displaystyle - 8r + 20 = 4r + 4$

    $\displaystyle - 8r -4r = 4 - 20$

    $\displaystyle -12r = -16$

    $\displaystyle r = \frac{-16}{-12}$

    $\displaystyle r = \frac{4}{3}$.

    The answer is that the bottom semicircle has a radius of $\displaystyle \frac{4}{3}$ cm.
    Last edited by johnsomeone; Oct 5th 2012 at 06:16 PM.
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  4. #4
    Senior Member MaxJasper's Avatar
    Aug 2012

    Lightbulb Re: Solve this please

    Radius of small tangent circle as a function of theta:

    Attached Thumbnails Attached Thumbnails Solve this please-tangent-circle-circle-inside-circle.png   Solve this please-tangent-circle-circle-inside-circle-r.png  
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