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Math Help - tangents and circles

  1. #1
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    tangents and circles

    Dear forum members,

    could someone please help me with this?

    Find the distance of the point P(the intersection of the tangents) from the centre of the smaller circle.

    The radius of the larger circle is 5, and the radius of the smaller is 3.That is all which is known.



    Could someone please give me a hint of how to start doing this?




    Thank you!
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  2. #2
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    Quote Originally Posted by Coach View Post
    Dear forum members,

    could someone please help me with this?

    Find the distance of the point P(the intersection of the tangents) from the centre of the smaller circle.

    The radius of the larger circle is 5, and the radius of the smaller is 3.That is all which is known.



    Could someone please give me a hint of how to start doing this?




    Thank you!

    Where the point P?
    AB=11
    AC=5
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  3. #3
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    Quote Originally Posted by SengNee View Post
    Where the point P?
    AB=11
    AC=5


    Both lines are tangents, and the point P is the intersection of the tangents.
    Attached Thumbnails Attached Thumbnails tangents and circles-tangent.jpg  
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  4. #4
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    Quote Originally Posted by Coach View Post
    Dear forum members,

    could someone please help me with this?

    Find the distance of the point P(the intersection of the tangents) from the centre of the smaller circle.

    The radius of the larger circle is 5, and the radius of the smaller is 3.That is all which is known.



    Could someone please give me a hint of how to start doing this?




    Thank you!
    There's almost certainly a simple circle geometry theorem I've forgotten that will make this question easy and the solution elegant.

    So in advance, here's a hamfisted solution outline (I'm assuming the two circles touch ...?) that might get you by until someone proves me right:

    Let C1 be the centre of the r = 5cm circle (hereby called circle 1) and let T1 be where the tangent touches this circle. Let C2 be the centre of the r = 3 cm circle (hereby called circle 2) and let T2 be where the tangent touches this circle.

    Use Pythag on triangle T1 C1 C2 to get T1 C2.
    Use pythag on triangle T1 T2 C2 to get T1 T2. I can't be bothered working it out, so I'll call the value a.

    Let the two tangents meet at P.
    Let T2 P = b and C2 P = c.
    Use Pythag on triangle C1 T1 P: 5^2 + (a + b)^2 = (8 + c)^2 ..... (1)
    Use Pythag on triangle C2 T2 P: 3^2 + b^2 = c^2 .... (2)

    Solve (1) and (2) simultaneously for c (remember that you know the value of a).
    Last edited by mr fantastic; January 13th 2008 at 05:03 AM. Reason: Disregard this method - it's rubbish.
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  5. #5
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    Quote Originally Posted by mr fantastic View Post
    There's almost certainly a simple circle geometry theorem I've forgotten that will make this question easy and the solution elegant.

    So in advance, here's a hamfisted solution outline (I'm assuming the two circles touch ...?) that might get you by until someone proves me right:

    Let C1 be the centre of the r = 5cm circle (hereby called circle 1) and let T1 be where the tangent touches this circle. Let C2 be the centre of the r = 3 cm circle (hereby called circle 2) and let T2 be where the tangent touches this circle.

    Use Pythag on triangle T1 C1 C2 to get T1 C2.
    Use pythag on triangle T1 T2 C2 to get T1 T2. I can't be bothered working it out, so I'll call the value a.

    Let the two tangents meet at P.
    Let T2 P = b and C2 P = c.
    Use Pythag on triangle C1 T1 P: 5^2 + (a + b)^2 = (8 + c)^2 ..... (1)
    Use Pythag on triangle C2 T2 P: 3^2 + b^2 = c^2 .... (2)

    Solve (1) and (2) simultaneously for c (remember that you know the value of a).
    Alternatively and a bit more simply:

    Consider trapezium T1 T2 C1 C2 and get angle C1 C2 T2. Then angle T2 C2 P is 180 minus angle C1 C2 T2. Now do trig on triangle T2 C2 P to get C2 P.
    Last edited by mr fantastic; January 13th 2008 at 05:05 AM. Reason: Simple AND correct, unlike my first suggestion
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    In your first method, I don't think Pythagoras is valid, because the diagram in my book does not state that the angle at C1 would be 90 degrees, and I don't know how valid an assumption as such would be. The place where the tangent touches the circles is perpedicular to the radius drawn to that point.




    The only theoretical thing I was taught that could be applied to this, is that the angle at C1+ angle at P=180
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  7. #7
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    Quote Originally Posted by Coach View Post
    In your first method, I don't think Pythagoras is valid, because the diagram in my book does not state that the angle at C1 would be 90 degrees, and I don't know how valid an assumption as such would be. The place where the tangent touches the circles is perpedicular to the radius drawn to that point.




    The only theoretical thing I was taught that could be applied to this, is that the angle at C1+ angle at P=180
    But note that the angles at T1 and T2 are each 90 degrees .....
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  8. #8
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    Use Pythag on triangle T1 C1 C2 to get T1 C2.


    True the angle at T1, is 90 degrees, but if I draw a hypotenuse from T1 to C2 then it will no longer be ninety. Then for this triangle T1 C1 C2 to become a right triangle, I would have to assume that the angle at C1 is 90 degrees, or am I just way off track here?




    if I use the first method(since I don't know how to use the second one), I get the following equation

    c^{2}+16c+64-25=80+2\sqrt{80}*\sqrt{c^2-9}



    and I don't know how to go on solving it from here.
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  9. #9
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    Quote Originally Posted by Coach View Post
    Use Pythag on triangle T1 C1 C2 to get T1 C2.


    True the angle at T1, is 90 degrees, but if I draw a hypotenuse from T1 to C2 then it will no longer be ninety. Then for this triangle T1 C1 C2 to become a right triangle, I would have to assume that the angle at C1 is 90 degrees, or am I just way off track here? Mr F says: Quite right. mea culpa.




    if I use the first method(since I don't know how to use the second one), I get the following equation

    c^{2}+16c+64-25=80+2\sqrt{80}*\sqrt{c^2-9}
    Mr F says: The first method is no good. See above.


    and I don't know how to go on solving it from here.
    Use the second method (it's easier and it's correct). Can you draw the trapezium?
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  10. #10
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    yes I think I can draw it.


    I know that on side is 8 units, and the two other sides are 3 and 5, and that thw two angles are 90 degrees. But how will that help me to figure out the other two angles? The sum of all the angles is 360?
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  11. #11
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    Quote Originally Posted by Coach View Post
    yes I think I can draw it.


    I know that on side is 8 units, and the two other sides are 3 and 5, and that thw two angles are 90 degrees. But how will that help me to figure out the other two angles? The sum of all the angles is 360?
    The trapezium diagram is attached. Do trig on the right-triangle C1 C2 A to get angle C1 C2 A. Angle C1 C2 T2 = 90 + angle C1 C2 A.

    Angle T2 C2 P = 180 - angle C1 C2 T2. Then do trig on triangle T2 C2 P to get C2 P.

    There a probably cleverer ways, but in the meantime see how you go.

    Whoops, forgot tto attach the trapezium - see next post.
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  12. #12
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    Quote Originally Posted by mr fantastic View Post
    The trapezium diagram is attached. Do trig on the right-triangle C1 C2 A to get angle C1 C2 A. Angle C1 C2 T2 = 90 + angle C1 C2 A.

    Angle T2 C2 P = 180 - angle C1 C2 T2. Then do trig on triangle T2 C2 P to get C2 P.

    There a probably cleverer ways, but in the meantime see how you go.

    Whoops, forgot tto attach the trapezium - see next post.
    ..
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