Results 1 to 3 of 3

Math Help - Coordinate geometry

  1. #1
    Member
    Joined
    Jul 2007
    Posts
    147

    Coordinate geometry

    Dear forum members,


    could you please help me?


    Find the values of b for which the straight line y=-x+b and the circle x^2+y^2-2x-2y-0.25=0 have no common points.


    So the distance from the line to the centre of the circle has to be more than the radius, right?


    I solved and obtained b<2-\frac{3}{2}*\sqrt{2}



    but the sample answer says:



    b<2-\frac{3}{2}*\sqrt{2}
    b>2+\frac{3}{2}*\sqrt{2}



    Why? Because of the square root?






    Thank you in advance!
    Follow Math Help Forum on Facebook and Google+

  2. #2
    Super Member wingless's Avatar
    Joined
    Dec 2007
    From
    Istanbul
    Posts
    585
    When you get \sqrt{ {\left ( \frac{b}{2} -1 \right )}^2 + {\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2},

    Notice that \sqrt{x^2} is NOT x, but it's |x|.

    \sqrt{ {\left ( \frac{b}{2} -1 \right )}^2 + {\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2}

    \sqrt{ 2\cdot{\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2}

    \sqrt{2}\cdot\sqrt{\cdot{\left ( \frac{b}{2} -1 \right )}^2 } < \frac{3}{2}

    \sqrt{2} \left | \frac{b}{2} -1 \right | < \frac{3}{2}

    Now solving this gives the correct answer.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Super Member

    Joined
    May 2006
    From
    Lexington, MA (USA)
    Posts
    11,546
    Thanks
    539
    Hello, Coach!

    Find the values of b for which the straight line y\:=\:-x+b
    and the circle x^2+y^2-2x-2y-0.25\:=\:0 have no common points.


    So the distance from the line to the centre of the circle
    has to be more than the radius, right? . . . . Right!
    As you found, the center is (1,1) and the radius is \frac{3}{2}

    I used this distance formula . . .

    The distance from a point (x_1,y_1) to the line ax + by + c \:=\:0

    . . is given by: . d \;=\;\frac{|ax_1 + by_1 + c|}{\sqrt{a^2+b^2}} . .
    Note the absolute value.


    We have point (1,1) and line x + y - b \:=\:0

    The distance is: . d \;=\;\frac{|1 + 1 - b|}{\sqrt{1^2+1^2}} . . . which is to be greater than \frac{3}{2}

    So we have: . \frac{|2-b|}{\sqrt{2}} \:>\:\frac{3}{2}\quad\Rightarrow\quad |2 - b| \:> \:\frac{3\sqrt{2}}{2}


    And we have two inequalities to solve . . .

    . . 2 - b \:>\:\frac{3\sqrt{2}}{2}\quad\Rightarrow\quad \text{-}b \:> \:\text{-}2 + \frac{3\sqrt{2}}{2} \quad\Rightarrow\quad\boxed{ b \:< \:2 - \frac{3\sqrt{2}}{2}}

    . . 2-b \:< \:\text{-}\frac{3\sqrt{2}}{2}\quad\Rightarrow\quad \text{-}b \:< \:\text{-}2 - \frac{3\sqrt{2}}{2} \quad\Rightarrow\quad \boxed{b \:> \:2 + \frac{3\sqrt{2}}{2}}

    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Coordinate Geometry
    Posted in the Geometry Forum
    Replies: 1
    Last Post: February 15th 2011, 11:06 PM
  2. Coordinate Geometry
    Posted in the Geometry Forum
    Replies: 1
    Last Post: June 5th 2010, 04:21 AM
  3. coordinate geometry
    Posted in the Algebra Forum
    Replies: 6
    Last Post: November 25th 2009, 07:28 AM
  4. coordinate geometry
    Posted in the Pre-Calculus Forum
    Replies: 3
    Last Post: November 24th 2009, 09:35 AM
  5. Coordinate Geometry
    Posted in the Pre-Calculus Forum
    Replies: 1
    Last Post: September 12th 2008, 04:26 PM

Search Tags


/mathhelpforum @mathhelpforum