Results 1 to 5 of 5

Math Help - more circle problems

  1. #1
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64

    more circle problems

    find the equation of circle with center on line y=3x and tangent to line x=2y at point (2,1).

    I graphed the line for the circle, I graphed the tangent line. I cant tell how to find a point on the circle line thats distance from (2,1) is going to be less then the distance from that point to any other point on the tangent line.

    I know (x,3x) is going to be the center. But how do I know that (x,3x) to (2,1) is more then (x,3x) to any other point on x=2y or (x,x/2).

    I tried putting x,3x as the equation of a circle from 2,1 :
    (2-x)^2+(1-3x)=r^2 and putting x=2y into this circle equation since x=2y == (x,x/2) and got (2-x)^2 + (1-3(x/2))^2 = r^2 but I kept going with this and got nowhere.

    Im probably missing something. can someone please point me in the right direction?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor
    Prove It's Avatar
    Joined
    Aug 2008
    Posts
    11,672
    Thanks
    1491
    A tangent to a circle is always perpendicular to the circle's radius.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    MHF Contributor

    Joined
    Apr 2005
    Posts
    15,987
    Thanks
    1650
    Here's one way to do it. A radius is always perpendicular to a tangent so a line through the center of the circle and the point of tangency is perpendicular to the tangent line. The line x= 2y has slope 1/2 and so any perpendicular line has slope -2.

    The line through (2, 1) with slope -2 has equation y= -2(x- 2)+ 1= -2x+ 4. The point where that line intersects y= 3x is the center of the circle and, of course, the distance from that point to (2, 1) is the radius.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Junior Member
    Joined
    Nov 2010
    From
    NY
    Posts
    64
    thanks people, never knew that the tangent line had to be perpendicular to the center of the circle. But that made it easy. thanks!
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,807
    Thanks
    697
    Hello, frankinaround!

    Another approach . . .


    \text{Find the equation of circle with center on line }y\,=\,3x
    \text{and tangent to line }x\,=\,2y\text{ at point }(2,1).

    The distance from (2,1) to the line 3x - y \:=\:0

    . . is given by: . d \:=\:\dfrac{|3(2) - 1(1)|}{\sqrt{3^2+1^2}} \;=\;\dfrac{5}{\sqrt{10}} \:=\:\sqrt{\dfrac{5}{2}}\quad \hdots\;\text{This is the radius..}


    The point (x,\,3x) which is \sqrt{\frac{5}{2}} units from (2,1)

    . . is given by: . \sqrt{(x-2)^2 + (3x-1)^2} \:=\:\sqrt{\frac{5}{2}}


    Square: . x^2 - 4x + 4 + 9x^2 - 6x + 1 \:=\:\frac{5}{2}

    . . which simplifies to: . 4x^2 - 4x + 1 \:=\:0

    . . which factors: . (2x-1)^2 \:=\:0

    . . and has one root: . x \,=\,\frac{1}{2}

    Hence: . y \:=\:3(\frac{1}{2}) \:=\:\frac{3}{2}


    The center is \left(\frac{1}{2},\,\frac{3}{2}\right) and the radius is \sqrt{\frac{5}{2}}

    The equation of the circle is: . \left(x - \frac{1}{2}\right)^2 + \left(y - \frac{3}{2}\right)^2 \;=\;\frac{5}{2}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. circle problems need help
    Posted in the Geometry Forum
    Replies: 2
    Last Post: May 17th 2010, 06:03 PM
  2. Circle Problems
    Posted in the Calculus Forum
    Replies: 3
    Last Post: June 24th 2009, 08:16 PM
  3. Circle geometry -2 problems
    Posted in the Geometry Forum
    Replies: 2
    Last Post: December 3rd 2008, 11:41 AM
  4. Ugh! Circle problems
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: October 10th 2008, 06:03 AM
  5. problems about further coordinate geometry(circle)
    Posted in the Pre-Calculus Forum
    Replies: 2
    Last Post: June 30th 2008, 12:19 AM

Search Tags


/mathhelpforum @mathhelpforum