Results 1 to 3 of 3

Math Help - Cirlce and line problem

  1. #1
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92

    Thumbs down Cirlce and line problem

    new to cirlces and getting stumped at this question. I'll run through my workings and thoughts - any advise would be greatly received.

    The equation of a cirlce with centre C is:  (x-3)^2+ (y-4)^2 = 9 and O is the origin. the line joining O and C can be extended to meet the cirlce at P. Find the coordinates of P and show the equation of the tangent to the cirlce at P is 3x + 4y = 40.

    So, I can observe the coords at C to be (3,4) and the line OP to have an equation of  y-0 = \frac{4}{3}(x-0) and therefore, y=\fra{4}{3}x.

    Substitute this into the circle equation and I get 75x^2 + 576x -108 = 0 which solves to be approx. x = 0.1831 and -7.8631

    But these x coords cant be right as I know the circle lies in the 1st quadrant and there for cannot have an intercept with the line at x = -7.8....

    What has gone wrong? Once I get the coords for the point P I can calculate the new equation of the line.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,660
    Thanks
    600
    Hello, dojo!

    Your algebra went a little off . . .


    \text{The equation of a circle with centre }\,C\text{  is:}
    . . (x-3)^2+ (y-4)^2 \:=\: 9\:\text{ and }O\text{ is the origin.}

    \text{The line joining }O\text{ and }C\text{ is extended to meet the circle at }P.

    \text{Find the coordinates of }P.

    \text{Show the equation of the tangent to the circle at }P\text{ is: }3x + 4y \:=\: 40


    \text{So, I can observe the coords at }C\text{ to be }(3,4) .Yes!
    \text{and the line }OP\text{ to have an equation: }\:y \:=\:\frac{4}{3}x .Right!

    \text{Substitute this into the circle equation}
    . . \text{and I get: }\:75x^2 + 576x -108 \:=\: 0 .no

    Substitute: . (x-3)^2 + \left(\frac{4}{3}x-4\right)^2 \:=\:9

    . . x^2 - 6x + 9 + \frac{16}{9}x^2 - \frac{32}{3}x + 16 \:=\:0

    . . . . . . . . . . . . \frac{25}{9}x^2 - \frac{50}{3}x + 16 \:=\:0

    Multiply by 9: . 25x^2 - 150x + 144 \:=\:0

    Factor:. . . . . . (5x - 6)(5x-24) \:=\:0

    . . Hence: . x \;=\;\frac{6}{5},\:\frac{24}{5}

    Since \,P is further from \,O. we have: . \boxed{P\left(\tfrac{24}{5},\:\tfrac{32}{5}\right)  }



    Line \,OP has slope \frac{4}{3}

    The tangent at \,P is perpendicular to OP.
    . . Hence, its slope is -\frac{3}{4}


    The line through \left(\frac{24}{5},\:\frac{32}{5}\right) with slope -\frac{3}{4} is:

    . . . . . . y - \frac{32}{5} \:=\:-\frac{3}{4}\left(x - \frac{24}{5}\right)


    Simplify: . y - \frac{32}{5} \:=\:-\frac{3}{4}x + \frac{18}{5}

    Multiply by 20: . 20y - 128 \:=\:-15x + 72 \quad\Rightarrow\quad 15x + 20y \:=\:200

    Divide by 5: . \boxed{3x + 4y \:=\:40}
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Member
    Joined
    Jan 2009
    From
    London
    Posts
    92
    LOL. I was applying the substitution to another circle expression in the question above! Thanks for pointing this out...
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Circle-Cirlce tangent in three dimensions
    Posted in the Geometry Forum
    Replies: 0
    Last Post: August 18th 2010, 02:14 PM
  2. Arc of an Cirlce
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 28th 2010, 02:44 AM
  3. Line problem
    Posted in the Geometry Forum
    Replies: 0
    Last Post: November 2nd 2009, 12:24 PM
  4. Replies: 0
    Last Post: October 22nd 2009, 11:04 AM
  5. Everone loves a cirlce(Worried)
    Posted in the Pre-Calculus Forum
    Replies: 5
    Last Post: March 31st 2008, 04:11 AM

/mathhelpforum @mathhelpforum