Results 1 to 2 of 2

Math Help - Translation of axes

  1. #1
    Member
    Joined
    Jan 2009
    Posts
    197

    Translation of axes

    Find the new equation of the circle of the equation  x^2+y^2-4x+6y+9=0 after the translation that moves the origin to the point (2,-3)

    Attempt
    x=x-2
    y=y+3

     (x-2)^2 +(y+3)^2 -4(x-2)+6(y+3)+9=0
     x^2+y^2-8x+12y+48=0

    Am I correct?
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,806
    Thanks
    697
    Hello, mj.alawami!

    Find the new equation of the circle of the equation  x^2+y^2-4x+6y+9\:=\:0
    after the translation that moves the origin to the point (2,-3).

    Attempt: . \begin{array}{c}x\:=\:x-2 \\ y\:=\:y+3 \end{array}

     (x-2)^2 +(y+3)^2 -4(x-2)+6(y+3)+9\:=\:0

     x^2+y^2-8x+12y+48\:=\:0

    Am I correct?

    Yes! . . . Good work!


    \begin{array}{cccccc}\text{The original circle is:} & (x-2)^2 + (y+2)^2 \:=\:4 &\Rightarrow& \text{Center: }(2,\,\text{-}3),\;r = 2 \\ \\[-3mm]<br />
\text{The new circle is:} & (x-4)^2 + (y+6)^2 \:=\: 4 &\Rightarrow& \text{Center: }(4,\,\text{-}6),\;r = 2 \end{array}

    And this checks out . . .

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. translation of axes
    Posted in the Geometry Forum
    Replies: 1
    Last Post: November 29th 2010, 09:03 PM
  2. Translation of axes
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 22nd 2009, 06:49 AM
  3. rotation of axes
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: May 11th 2009, 07:37 PM
  4. Semi-axes of an ellipse
    Posted in the Advanced Math Topics Forum
    Replies: 0
    Last Post: October 25th 2007, 10:47 AM
  5. ellipse axes
    Posted in the Calculus Forum
    Replies: 1
    Last Post: October 19th 2006, 04:16 AM

Search Tags


/mathhelpforum @mathhelpforum