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Equation for the tangent of a circle

  1. Apr 4, 2010 #1
    1. The problem statement, all variables and given/known data
    If the equation of one tangent to the circle with center at (2, -1) from the origin is 3x + y = 0, then the equation of the other tangent through the origin is:
    (a) 3x - y = 0
    (b) x + 3y = 0
    (c) x - 3y = 0
    (d) x + 2y = 0


    2. Relevant equations

    An equation of the tangent to the circle [tex]x^2 +y^2 + 2gx + 2fy + c = 0[/tex] at the point [tex]( x_{1}, y_{1})[/tex] on the circle is

    xx1 + yy1 + g(x + x1) + f (y + y1) + c = 0

    There are more but I presume that you know

    3. The attempt at a solution

    I've already solved this question from another method but just asking out of my curiosity.

    I've learned it on my lower classes that radius are perpendicular to the tangent of the circle. It means that we must get m1. m2 = -1 but as you can see we are not getting, neither in the case of first tangent nor in the case of second (in any of the option given). Can you tell me, why?
     
  2. jcsd
  3. Apr 4, 2010 #2
    nobody is answering. I thing I've asked either a very easy question or a very silly one.
     
  4. Apr 4, 2010 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    No, you just don't have patience. Your second post came 43 minutes after the first. You can't expect people to be sitting at their computers waiting for you to ask a question!

    One way to do this is to note that the line 3x+ y= 0 has slope -3 while the line from the origin to (2, -1) has slope -1/2. If we write the angle the first line makes with the x-axis as [itex]\theta_2[/itex] then [itex]tan(\theta_2)= -3[/itex]. If we write the angle the line from the origin to (2, -1) as [itex]\theta_0[/itex] then we have [itex]tan(\theta_0)= -1/2[/itex]. Let [itex]\theta_1[/itex] be the angle the other tangent makes with the x-axis, then we have
    [tex]\theta_0= \frac{\theta_1+ \theta_2}{2}[/tex]
    so that [itex]\theta_1= 2\theta_0- \theta_2[/itex].

    Now use some trig identities, like
    [tex]tan(a- b)= \frac{tan(a)- tan(b)}{1+ tan(a)tan(b)}[/tex]
    and
    [tex]tan(2a)= \frac{2 tan(a)}{1- tan^2(a)}[/tex]

    to find [itex]tan(\theta_1)[/itex], the slope of the second tangent line.
     
  5. Apr 4, 2010 #4
    thanks but in our country (India) we have to solve such questions in less than 1 minute so I can't go this way. There must be some easy method. Do you know any trick or any more concept to solve it with less calculation involved. If this question is given in our book then there must be some easy method to solve it.

    We get objective type questions and a very less time (less than 1 minute) for every questions.
     
    Last edited: Apr 4, 2010
  6. Apr 4, 2010 #5
    I think I'm very close to get answer but I want to know what is the angle between two tangents drawn from the same point on a circle.
     
  7. Apr 4, 2010 #6
    Oh! Sorry. I'd read that question wrong. It is a very simple question.

    I thought that origin is on circle where tangent touch the circle but now after reading the question again I've got that the origin is not on circle but outside of circle from where two tangents are drawn.

    Thanks HallsofIvy for your help. I've got a very easy solution which can be done in a minute.
     
  8. Apr 4, 2010 #7
    What I've done is I've found the length of radius of circle (by using the formula of length of perpendicular from a point) and then I've assumed the equation of other tangent as y = mx and again calculated the length of radius in terms of m (slope) and then equated it with the length which I've got previously. Then I've got two values of m (for both the tangent). One was -3 and other was 1/3 so the equation of other tangent I've found is x - 3y = 0

    do you know any better solution than this. If You know then please tell me. I've my exam on 11th of April.
     
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