Results 1 to 7 of 7

Math Help - circles

  1. #1
    Member
    Joined
    Aug 2009
    Posts
    96

    circles

    O is the centre of the small circle. CA and CB are tangents to this circle. The large circle is drawn through A, B and C. CO is produced to E and intersects the small circle at D.
    i. Show that O lies on the large circle

    ii. Show that CD = (CB x CA) / (CD + 2DO)
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member
    Joined
    Jan 2009
    Posts
    591
    Quote Originally Posted by deej813 View Post
    O is the centre of the small circle. CA and CB are tangents to this circle. The large circle is drawn through A, B and C. CO is produced to E and intersects the small circle at D.
    i. Show that O lies on the large circle

    ii. Show that CD = (CB x CA) / (CD + 2DO)
    I can't do it!.
    Either there is something missing or mistyped.
    The circumcircle ABC completely encloses the small circle with center at O
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Newbie
    Joined
    Aug 2009
    Posts
    14
    Quote Originally Posted by aidan View Post
    I can't do it!.
    Either there is something missing or mistyped.
    The circumcircle ABC completely encloses the small circle with center at O
    Is it not something like this drawing below.
    (D & E might be the wrong way round)
    Mind you, even with the drawing I cant answer the OPs question,
    I can see that no 1 is true but I have no idea how to prove it.
    Attached Thumbnails Attached Thumbnails circles-madudanracun.png  
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Super Member Matt Westwood's Avatar
    Joined
    Jul 2008
    From
    Reading, UK
    Posts
    824
    Thanks
    33
    Let Q be the point on CO at which the larger circle intersects CO. We want to show that this O.

    Since CQ is a diameter of the larger circle, angle CAQ is a right angle (standard theorem in geometry). So as CA is a tangent of the smaller circle, and is perpendicular to AQ, it follows that AQ must be a radius of the smaller circle and so AQ = AO.

    So Q = O.

    That's the first bit done.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by deej813 View Post
    O is the centre of the small circle. CA and CB are tangents to this circle. The large circle is drawn through A, B and C. CO is produced to E and intersects the small circle at D.
    i. Show that O lies on the large circle

    ii. Show that CD = (CB x CA) / (CD + 2DO)
    I think the picture should look like the attachment. Follow Matt Westwood's advice for part (i). For (ii), notice that CB = CA and CE = CD + 2DO (assuming that E is the point where the line CDO crosses the small circle again). Then the result CD = (CB x CA) / (CD + 2DO) is equivalent to CD\times CE = CA^2. This is a standard result which you can prove for example by using the fact that the triangles ADC and EAC are similar.
    Attached Thumbnails Attached Thumbnails circles-circles.jpg  
    Follow Math Help Forum on Facebook and Google+

  6. #6
    Member
    Joined
    Aug 2009
    Posts
    96
    ok thanks i get i.
    but with ii. i dont get how CD = (CB x CA) / (CD + 2DO) is equivalent to CD x CE = CA^2
    Follow Math Help Forum on Facebook and Google+

  7. #7
    MHF Contributor
    Opalg's Avatar
    Joined
    Aug 2007
    From
    Leeds, UK
    Posts
    4,041
    Thanks
    7
    Quote Originally Posted by deej813 View Post
    ok thanks i get i.
    but with ii. i dont get how CD = (CB x CA) / (CD + 2DO) is equivalent to CD x CE = CA^2
    Multiply out the fraction. Then the equation CD = (CB x CA) / (CD + 2DO) becomes CD (CD + 2DO) = CB CA. But you can see from the picture that CE = CD + DO + OE, where DO and OE are both equal to the radius of the small circle. So CE = CD + 2DO. Also, CB = CA (because the two tangents from a point to a circle are always equal).
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Prove that every rigid motion transforms circles into circles
    Posted in the Differential Geometry Forum
    Replies: 4
    Last Post: February 11th 2010, 06:00 PM
  2. Replies: 2
    Last Post: October 6th 2009, 08:04 AM
  3. going in circles
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: July 23rd 2009, 10:57 AM
  4. Circles!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 16th 2009, 07:50 PM
  5. circles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: October 16th 2008, 03:40 AM

Search Tags


/mathhelpforum @mathhelpforum