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Math Help - Circles and Internal Tangent Segments

  1. #1
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    Thumbs up Circles and Internal Tangent Segments

    Hi Forum.
    I have been studying secant and tangent theorems for the circles and I found this question:

    Find the length of the common internal tangent segment of two circles with radii of 4 units and 12 units whose centers are 20 units apart.

    Ok, I drew a sketch and I tried to used secant tangent theorems but I didn't get it off the ground.
    There's a formula that states x=\sqrt{y^2-(r_1+r_2)^2}
    With y being the segment's length.
    I can only think of ways of solving this, cutting the segment into two parts, that are not equal, and then summing them.
    But I don't know how to put the distance of 20 units between the centers.

    My question is, how do we get the formula x=\sqrt{y^2-(r_1+r_2)^2}?

    Thanks!

    Note that the x and the y in the formula are not the same x and y in the sketch.
    Here is a sketch:
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  2. #2
    MHF Contributor Also sprach Zarathustra's Avatar
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    Re: Circles and Internal Tangent Segments

    Quote Originally Posted by Zellator View Post
    Hi Forum.
    I have been studying secant and tangent theorems for the circles and I found this question:

    Find the length of the common internal tangent segment of two circles with radii of 4 units and 12 units whose centers are 20 units apart.

    Ok, I drew a sketch and I tried to used secant tangent theorems but I didn't get it off the ground.
    There's a formula that states x=\sqrt{y^2-(r_1+r_2)^2}
    With y being the segment's length.
    I can only think of ways of solving this, cutting the segment into two parts, that are not equal, and then summing them.
    But I don't know how to put the distance of 20 units between the centers.

    My question is, how do we get the formula x=\sqrt{y^2-(r_1+r_2)^2}?

    Thanks!

    Note that the x and the y in the formula are not the same x and y in the sketch.
    Here is a sketch:
    From Pythagoras theorem:

    (1) 12^2+t_2^2 =(x+12)^2

    (2) 4^2+t_1^2=(y+4)^2

    Now, you now that (3) x+y=4 (why?)

    And also you have two similar triangles, so:

    (4) \frac{t_1}{t_2}=\frac{4}{12}

    All the above (1), (2), (3) and (4) should be enough for you solving the problem...


    EDIT:

    You can arrive to your formula by 'playing' with the above equations...(try it by your own)
    Last edited by Also sprach Zarathustra; June 17th 2011 at 06:15 PM.
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  3. #3
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    Re: Circles and Internal Tangent Segments

    Quote Originally Posted by Zellator View Post
    Hi Forum.
    I have been studying secant and tangent theorems for the circles and I found this question:

    Find the length of the common internal tangent segment of two circles with radii of 4 units and 12 units whose centers are 20 units apart.

    ...
    1. I've modified your sketch a little bit (see attachment).

    2. Use proportions with the similar (greyed) triangles:

    \dfrac{t_1}{t_2}=\dfrac{r_1}{r_2}~\implies~t_1 = \frac{r_1}{r_2} \cdot t_2

    3. Since you know the distance between the centers and the lengthes of the radii you can use Pythagorean theorem to determine t_1+t_2 = 12.

    4. Replace t_1 in the last equation by the term of #2.
    Solve for t_2 and afterwards for t_1.
    Attached Thumbnails Attached Thumbnails Circles and Internal Tangent Segments-interntangsegm.png  
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  4. #4
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    Re: Circles and Internal Tangent Segments

    Hi Zellator,
    You can solve this problem using your sketch.

    x+y = 4
    4/12 =y+4/x+12
    whence x= 3 y=1
    (12+x)^2 =4^2 +(t2)^2
    15^2 = 4^2 + (t2)^2
    t2 =9
    t1/t2 = 1/3
    t1 =3




    bjh
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  5. #5
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    Re: Circles and Internal Tangent Segments

    Hey Guys!
    Thanks for your replies! That's what I love about this forums!
    Thanks for your clever answers Also sprach Zarathustra, earboth and bjhopper!
    I did considered them all, it's great to have this alternative ways to do it.
    Like Also sprach Zarathustra said, you can play with the possibilities here.

    Thanks again!
    All the best!
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