Results 1 to 5 of 5

Math Help - Circles

  1. #1
    Newbie
    Joined
    Nov 2008
    Posts
    21

    Exclamation Circles

    Problem: A circle A passing through the point (4, 0) is orthogonal to a circle B with equation x^2+y^2=4. Show that the locus of the centre of the circle A is a straight line and find its equation.

    Well, I first established the centre and radius of circle B as (0, 0) and 2 respectively. Then, I let the centre of circle A be (a,b). Then, I reasoned that since circles are orhtogonal, the square of the distance between the centre should be equal to the sum of the squares of the two radii, so that: a^2+b^2 = 4 + (0-a)^2 + (4-b)^2. This gave me the value of a but then I got stuck. Was I on the right track? Any suggestions please?

    Thanks in advance
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Junior Member
    Joined
    Apr 2008
    Posts
    51
    This isn't about your question but try using the math tab to make things less confusing. Looks like a fancy E. x^2+y^2=4

    See?
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by yobacul View Post
    Problem: A circle A passing through the point F(4, 0) is orthogonal to a circle B with equation x^2+y^2=4. Show that the locus of the centre of the circle A is a straight line and find its equation.

    ...
    1. If A \perp B~\implies~ radius of A must be tangent to B that means A \perp B~\implies~r_A \perp r_B

    2. If C(a, b) as you suggested then C is the point of intersection between the tangent in T and the perpendicular bisector of TF.

    3. I've attached a drawing of the situation. The locus of all C is drawn in red.

    EDIT: I used a slightly different approach

    Let T(t, \sqrt{4-t^2})

    Then the tangent in T to B has the equation:

    y = \dfrac{4-tx}{\sqrt{4-t^2}} and the perpendicular bisector of TF has the equation y = -\dfrac{x(4-t)+6}{\sqrt{4-t^2}}

    Calculate the x-coordinate of the point of intersection between these two lines, that means solve for x:

    \dfrac{4-tx}{\sqrt{4-t^2}} = -\dfrac{x(4-t)+6}{\sqrt{4-t^2}} which yields x = \dfrac52 . As you can see there isn't a parameter t anymore. So x = 2.5 is the equation of the locus of the centers of all possible circles A.
    Attached Thumbnails Attached Thumbnails Circles-perpcirc_ortslinie.png  
    Last edited by earboth; October 5th 2009 at 11:38 AM. Reason: additional calculations
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Newbie
    Joined
    Nov 2008
    Posts
    21

    Exclamation Still confused ...

    Many thanks for answering the question. I tried to go through all the steps but I am still confused. I understood why T takes those coordinates but then I got lost.
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member
    earboth's Avatar
    Joined
    Jan 2006
    From
    Germany
    Posts
    5,830
    Thanks
    123
    Quote Originally Posted by yobacul View Post
    Many thanks for answering the question. I tried to go through all the steps but I am still confused. I understood why T takes those coordinates but then I got lost.
    I've found a much simpler way to solve this question:

    1. See attachment. Actually you are dealing with 3 right triangles, labeled I, II and III.

    2. Let r denote the radius of circle B and R denote the radius of circle A. C(a, b) is the center of circle A.

    3. Then you get a system of equations:

    \left|\begin{array}{rcl}r^2+R^2&=&a^2+b^2\\ R^2&=&(4-a)^2+b^2\end{array} ~\implies~\right. \left|\begin{array}{rcl}4+R^2&=&a^2+b^2\\ R^2&=&16-8a+a^2+b^2\end{array} \right.

    Now subtract columnwise:

    4 = -16 +8a~\implies~\boxed{a=\dfrac52}

    This result is valid for all circles A which satisfy the given conditions.
    Attached Thumbnails Attached Thumbnails Circles-perpcirc_locus.png  
    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, 07:00 PM
  2. Replies: 2
    Last Post: October 6th 2009, 09:04 AM
  3. going in circles
    Posted in the Pre-Calculus Forum
    Replies: 4
    Last Post: July 23rd 2009, 11:57 AM
  4. Circles!
    Posted in the Trigonometry Forum
    Replies: 5
    Last Post: July 16th 2009, 08:50 PM
  5. circles
    Posted in the Geometry Forum
    Replies: 1
    Last Post: October 16th 2008, 04:40 AM

Search Tags


/mathhelpforum @mathhelpforum