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Homework Help: Find the equation of the tangent

  1. Apr 12, 2008 #1
    1. The problem statement, all variables and given/known data

    Find the equation of the tangent of the circular [itex]K:x^2 + y^2 - 2x + 4y=0[/itex], perpendicular to the line x-2y+9=0.

    2. Relevant equations

    [itex](x_1-p)(x-p)+(y_1-q)(y-q)=r^2[/itex], equation of K.

    [itex](kp-q+n)^2=r^2(k^2+1)[/itex], condition for tangent and circular K

    3. The attempt at a solution

    I tried like this. From the equation [itex]K:x^2 + y^2 - 2x + 4y=0[/itex],
    p=1 and q=-2 ,[itex]r^2=5[/itex]
    also from x-2y+9=0, the coefficient k of the equation of the tangent should be k=-2.
    So I have y=-2x+n, and -2x-y+n=0

    from here I get [itex]n_1=-5[/itex], and [itex]n_2=5[/itex], so the equation should be:



    But the problem is that it is not same with my text book results. Any help?
  2. jcsd
  3. Apr 12, 2008 #2
    Solve the equation of the tangent in terms of k with the equation of the given line to obtain a third point .
  4. Apr 12, 2008 #3


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    Science Advisor

    Is [itex]x_1[/itex] different from x?

    Wouldn't it be simpler just to use basic concepts rather than such complicated formulas?

    The line x- 2y+ 9= 0 can be written 2y= x+ 9 or y= (1/2)x+ 9/2 so its slope is 1/2. Any line perpendicular to that must have slope -2. For the given circle, [itex]x^2 + y^2 - 2x + 4y=0[/itex], implicit differentiation gives 2x+ 2yy'- 2+ 4y'= 0 or (2y+ 4)y'= 2- 2x so y'= (2- 2x)/(2y+ 4)= (1- x)/(y+ 2). For what values of x and y is that equal to 2?

    You have two equations: (1-x)/(y+2)= 2 and [itex]x^2 + y^2 - 2x + 4y=0[/itex] to solve. Solve the first for either x or y, put that into the second equation and solve the resulting quadratic.
  5. Apr 12, 2008 #4


    What to do now? x and y are different from x_1 and y_1
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