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Homework Help: Coordinate geometry of the circle question

  1. Feb 21, 2009 #1
    1. The problem statement, all variables and given/known data
    Fairly straight forward question but I just can't see whats wrong.

    A circle passes through the point (2,-4) and touches both the x-axis and the y-axis. Find the equations of the two circles which satisfy these conditions.

    2. Relevant equations

    [tex] x^2 + y^2 + 2gx + 2fy + c = 0[/tex]

    with a centre point [itex] c (-g,-f) [/itex]

    [tex] r = \sqrt{g^2 + f^2 - c} [/tex]

    Where r is the radius.
    3. The attempt at a solution

    After drawing a diagram I concluded that r=g and r=f therefore g=f.

    [tex] g = \sqrt{g^2 + f^2 - c} [/tex]

    [tex] g^2 = g^2 + f^2 - c [/tex]

    [tex] f^2 = c [/tex]

    Since the point (2,-4) is on the circle it will satisfy :

    [itex] x^2 + y^2 + 2gx + 2fy + c = 0[/itex]

    [tex] (2)^2 + (-4)^2 + 2g(2) + 2f(-4) + c = 0 [/tex]

    [tex] 4 + 16 + 4g -8f + c = 0 [/tex]

    [tex] 20 + 4g -8f + c = 0 [/tex]

    But [itex] f^2 = c [/itex] and [itex] f = g [/itex]


    [tex] 20 + 4f - 8f + f^2 = 0[/tex]
    [tex] 20 -4f + f^2 = 0 [/tex]
    [tex] f^2 -4f +20 = 0 [/tex]

    This quadratic only has complex roots.

    Thanks for any help!
  2. jcsd
  3. Feb 21, 2009 #2


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    If the center is [itex](x_0,y_0)[/itex], that is, the equation of the circle is [itex](x-x_0)^2+ (y-y_0)^2= R^2[/itex] and it "touches" the axes- by which I assume you mean it is tangent to them, then [itex](x_0,0)[/itex] and [itex](0,y_0)[/itex] are the points at which the circle is tangent to the axes. Putting [itex]x= x_0[/itex], y= 0 into the equation, [itex](-y_0)^2= R^2[/itex] and putting [itex]y= y_0[/itex], x= 0, [itex](-x_0)^2= R^2[/itex] so we must have [itex]x_0= y_0= R[/itex] so we can write the equation as [itex](x- R)^2+ (y- R)^2= R^2[/itex]. Adding the condition that (2,-4) lies on that circle gives [itex](2-R)^2+ (-4-R)^2= R^2[/itex] which you can solve for R.
  4. Feb 23, 2009 #3
    Yeah so you get [tex] R^2 + 4R + 20 = 0[/tex]

    this has complex roots as well....
  5. Feb 23, 2009 #4


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    The center of the circle isn't at (R,R). That's in the wrong quadrant. You have to pick y0 to be the negative root of y0^2=R^2. Put it at (R,-R).
  6. Feb 23, 2009 #5
    Ok cool thanks so it is

    [tex] R^2 -12R +20 = 0 [/tex]

    [tex] R=2 [/tex] and [tex] R=10 [/tex]

    Therefore the equations of the circles are

    [tex] (x- R)^2+ (y+ R)^2= R^2 [/tex]

    [tex] (x- 2)^2+ (y+ 2)^2= R^2 [/tex]


    [tex] (x- 10)^2+ (y+ 10)^2= R^2 [/tex]

    Just out of curiosity what was wrong with my original attempt?
  7. Feb 23, 2009 #6


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    Exactly the same thing. You said f=g=r. But that's not true. |f|=|g|=r is what is true. You have to figure out the signs based on the location of (2,-4).
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