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Well, I thought it was going to be easy.

  1. Sep 17, 2009 #1
    1. The problem statement, all variables and given/known data
    Find two equations for lines that pass through the point (5,6) and are tangent to 1=x^2+y^2.


    2. Relevant equations

    y=mx +b


    3. The attempt at a solution

    Well, I thought this was going to be an easy one but i'm running out of creativity. I would still like to solve this one so it would be nice if I could get a small shove in the right direction.

    I starting thinking about the properties of a unit circle in order to locate a point on the circle. I was then planning on taking the derivative at that point to find the slope of the line and construct the equation from there.

    The problem with that approach is that I am not sure how to figure out what point that is. I am open to trying another approach as well.
     
  2. jcsd
  3. Sep 17, 2009 #2
    If the line is tangent to x2+y2=1 at the point (5,6), then what do you know about the line's slope?
     
  4. Sep 17, 2009 #3

    Office_Shredder

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    Think about that one again.

    here's the situation
    http://img43.imageshack.us/img43/6778/twolines.jpg [Broken]

    I would use this theorem to solve it
    http://en.wikipedia.org/wiki/Power_of_a_point

    Since we can find how far the points of intersection of the circle are from (5,6) and hence we can solve for exactly which points they are (using as a reference the line passing from (5,6) directly through the center of the circle)
    But I'm not sure if that's what's intended
     
    Last edited by a moderator: May 4, 2017
  5. Sep 17, 2009 #4
    Ah ha, tricky. I misinterpreted the question :redface: Nice catch Office Shredder :smile:
     
    Last edited by a moderator: May 4, 2017
  6. Sep 17, 2009 #5
    I started to try to solve this myself and drew the same on paper.

    Thanks for posting the link with the method for solution.

    I couldn't figure it out either.
     
  7. Sep 17, 2009 #6

    Office_Shredder

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    I gave this some more thought and came up with this:

    A line passing through (5,6) has equation y=mx+b for unknown m and b. This gives us one equation to help solve for A lot of these lines pass through the circle. In fact, y=mx+b passes through the circle precisely if there exist x and y such that x2 + (mx+b)2 = 1 is solvable. (plugging y=mx+b into the circle equation)

    This equation is

    x2 + m2x2 + 2mxb + b2 = 1

    The line is tangent to the circle precisely when there is only one solution to this polynomial. Aka the discriminant is zero. This means that the line is tangent if and only if
    (2mb)2 -4(1+m2)(b2-1) = 0
    Using this and

    6=5m+b

    gives two equations in two unknowns. Not sure if you can solve it though (and I don't really want to try)
     
  8. Sep 18, 2009 #7
    Using y = √(1 - x2) for the tangent point (a, y(a)) in the second quadrant:

    y'(a) must be equal to the slope of the line which goes through the points (5, 6) and (a, y(a)).
    m = y'(a) and m = (6 - y(a))/(5 - a). There you get an equation with one variable and hopefully it works out nicely.
     
  9. Sep 18, 2009 #8

    Office_Shredder

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    Yeah, but you don't know what a or y(a) is, which is the difficult part
     
  10. Sep 18, 2009 #9
    Well, we just plug a into y (and y'): y(a) = √(1 - a2)
    y'(a) = -a/√(1 - a2)

    m = y'(a)
    m = (6 - y(a))/(5 - a)

    [tex]\frac{-a}{\sqrt{1 - a^2}} = \frac{6 - \sqrt{1 - a^2}}{5 - a}[/tex]

    Just one thing to solve for: a. Plug that into y and you have the coordinates of one point.
     
  11. Sep 18, 2009 #10
    Thank you to all who have responded. This was a bit more work than I had expected.

    Bohrok,

    Without using a computer, I am not sure how to solve for a. I must admit that I am embarrassed to be so rusty in my algebra. Could you (or anyone) provide some hints on what to do with the quantities sqrt(1-a^2)?
     
  12. Sep 18, 2009 #11

    sylas

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    Rearrange so that they are isolated on one side of the equation, then square both sides.
     
  13. Sep 18, 2009 #12
    "Find two equations for lines that pass through the point (5,6) and are tangent to 1=x^2+y^2."

    I would think one could rearrange the equation 1 = x^2+y^2, by solving y. Then use
    derivative rules to find an equation for a slop at point x. Then you have a point, (5,6),
    use that to find 1 equation. The other equation could probably be found from Saladsamurai
    picture. You can see that the other equation is at point (-5,-6), (not really sure though),
    So use that point to find the other equation.

    Am I missing something?

    [edit] That part that I was assuming,the second equation might be mirrored at point(-5,-6) could probably be shown by
    symmetry of a circle. [/edit]

    [edit = 2] I also see that when you get your mx+b equation, you have to make it a unit vector.
    so l = sqrt(m^2 + b^2), then now m = m/l and b = b/l which means y = mx/l + b/l, where l is the sqrt(m^2+b^2);
    [/edit = 2]
     
    Last edited: Sep 18, 2009
  14. Sep 19, 2009 #13

    sylas

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    You can do this all much more easily if you go back and write down all the equations.

    The original post listed one equation only

    [tex]y = mx + b[/tex]​

    That has four variables. Suppose you let "y" and "x" in this case denote one specific point on the line; where it touches the unit circle at a tangent. You also have the point (6,5) on the line defined by m and b and so this gives the relation the OfficeShredder provides in msg #6.

    [tex]6 = 5m + b[/tex]​

    OfficeShredder also derives a relation for m and b, but it is a quartic. We should be able to get a quadratic... and we can. Bohrok shows the way... but introduces variables "a" and "y(a)" to represent the tangent point. I've just kept with x and y for this point, so his equation becomes m=(6-y)/(5-x)

    But this is what you get by canceling b from the previous two equations, so it is not actually a new independent equation. We still need two more equations. One is the fact that the tangent point (x,y) is on the unit circle.
    [tex]1 = x^2 + y^2[/tex]​

    The last equation has to capture the information that it is on the tangent, not just anywhere on the circle.

    Bohrok has given y'(a) = -a/√(1 - a2), and this works. But let's keep it basic and not cancel variables until we've got all the equations. Remember that "a" is just x, and the bottom term is just y, and the gradient has to be m. We can just write Bohrok's equation as follows:
    [tex]m = -x/y[/tex]​

    A nice way to understand this last equation is to look at the gradient of the line from the origin to (x,y). This is, by definition, y/x. It is also at right angles to m, since it is a tangent, and that means m is the negative of the inverse; which is what the equation says.

    So. Here are four equations in four unknowns.
    [tex]\begin{align}
    y &= mx + b \\
    6 &= m5 + b \\
    1 &= x^2 + y^2 \\
    m &= -x/y
    \end{align}[/tex]​

    The question is asking for lines, and so what you really want is m and b. Also, in substituting variables, it will be best to avoid taking square roots if possible, so that we don't get messed up with signs. So use equations 1, 2 and 4, which are nice simple linear relations, to give substitutions. Get rid of x and y early, because it is the line you want.

    You should be able to get a simple quadratic equation for m out of equation (3), which is then easily solved for the two lines. The corresponding values for b will come from equation (2).

    Everyone; don't rush in with a solution too quickly. I may even have done a bit too much myself here; but I'm trying to pull together the various inputs into a simple and usable form.

    PhysicsMark... over to you again!

    Cheers -- sylas
     
    Last edited: Sep 19, 2009
  15. Sep 19, 2009 #14
    Consider one of the tangents and draw the following lines
    point (5,6) to centre of circle
    point(56) to tangent where it meets circle
    centre of circle to tangent point
    You now have a right angled triangle and you can take it from there
     
  16. Sep 19, 2009 #15

    sylas

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    Except that the gradient you want for the line is not any of the angles in that triangle. You can do it from there, but it will involve more triangles. In the end you should get equations similar to those that have been identified in the thread.
     
  17. Sep 19, 2009 #16
    I know the gradient is not one of those angles.I just thought it to be an easy way to start the problem.
     
  18. Sep 20, 2009 #17
    I believe a I have an idea similar to what Dadface said.

    Sylas,
    You seem pretty confident in your method. Would you look over this idea and tell me if it makes sense to you?

    Here is what I did:

    I drew a line from point (5,6) to the origin. I labeled the distance of this line to be sqrt(61). I then drew two perpendicular lines(to the slope of the first line) of length 2 at each end (one going through (0,0) and the other going through (5,6). At the endpoint of the perpendicular lines, I drew two lines each with slope 6/5 to connect the two perpendicular lines. (I apologize if this is confusing...I am not sure how to upload a picture)

    This gave me two rectangles with side sqrt(61) and side 1. I then drew a diagonal line from point (5,6) to unknown point (x,-y). I labeled this line length sqrt(62). I used the inverse cosine of (sqrt(61)/sqrt(62)) to find the angle that the diagonal line makes with the very first line I drew. This angle was about 7.2962 degrees. I next took the inverse cosine of 5/sqrt(61) to find the angle the first line that was drawn makes with the line parallel to y=0. I found this able to be about 50.1944. I added the two angles next. This left me with about 57.4906 degrees.

    Now that I had the degree the and the distance from (5,6) to the unknown point(Or so I think) I used the polar coordinate conversion to find that

    x = (5-sqrt(62)cos(57.4906) = .7682
    y = (6-sqrt(62)sin(360-57.4906) = -.640184

    I next used point (5,6) and point (.7682, -.640184) to construct and equation for the first line.

    Ok, So I came up with y=1.569124x-.845619

    I used so many figures because I wanted an accurate graph. I used my Ti-83 to graph the lower half of the circle and my line. It is not accurate when the graph is zoomed in on, but I am not sure if this is because of my rounding or my approach.

    Basically I am not sure how my answer stacks up. I only did the one line because I would like some feed back before I continue.

    Looking back at others' posts, I realize that my post was not very clear from start. I am sorry for the confusion this has caused. This problem was one I found online. After tinkering around and getting stumped I came here for some advice. I realize now that I should have not restricted the equations to just y=mx+b. I was thinking that that particular equation was going to be useful. I did not think about it being restricted as the only equation to use. Sorry fellas.
     
  19. Sep 20, 2009 #18

    sylas

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    Your rectangles have a vertex lying on the circle, but it is not at the tangent point. This leads to the wrong answer, though the difference is small.

    If you think about the rightangle triangle described by DadFace, the hypotenuse is of length sqrt(61), and so the length of the line from (5,6) to the tangent point will be sqrt(60), not sqrt(62). (I have attached a diagram.) The angle between the line (0,0)-(5,6) and the tangent lines is atan(1/sqrt(60)).
    tangents.JPG

    Your value, of 7.2962 degrees, is atan(1/sqrt(61)), or asin(1/sqrt(62)), but that's not the angle you want. Your value of 50.1944 degrees for the line from (0,0) to (5,6) is fine.

    You're close, but you've basically missed the tangent point.

    Try again, using Dadface's triangles rather than the rectangles. Everything else is okay. You'll get an angle something close to what you had before, a bit over 7 degrees.

    For comparison, here's what I got using my method.

    From equation (4) I know x = -my
    From equation (2) I know b = 6 - 5m

    Substitute for x in equation (3) gives m2y2 + y2 = 1

    Substitute for x and b into equation (1) gives y = 6 - 5m - m2y

    Getting a single y in each of these equations gives these two equations in y and m
    [tex]\begin{align*}
    y^2(1 + m^2) &= 1 \\
    y(1 + m^2) &= 6 - 5m
    \end{align*}[/tex]​

    Square both sides of the second equation, and then substitute the first equation for y2.
    [tex]\begin{align*}
    y^2(1 + m^2) &= 1 \\
    y^2(1 + m^2)^2 &= (6 - 5m)^2 \\
    (1 + m^2) &= (6 - 5m)^2 \\
    &= 36 - 60m + 25m^2 \\
    0 & = 35 - 60m + 24m^2
    \end{align*}[/tex]​

    This is now a quadratic equation for m, which has solution:
    [tex]\begin{align*}
    m &= \frac{60 \pm \sqrt{60^2 - 4 \times 35\times 24}}{48} \\
    &= \frac{15 \pm \sqrt{15^2 - 35 \times 6}}{12} \\
    &= \frac{15 \pm \sqrt{15}}{12} \\
    &= 0.927251388 \;\; \text{or} \;\; 1.572748612
    \end{align*}[/tex]​

    These are the gradients for the lines. The corresponding angles are obtained using atan(m), and this is 57.5506 degrees and 42.8383 degrees. If you fix up your method, you should get these numbers as well (you add and you subtract from the 50.1944 degrees.)

    Cheers -- sylas
     
    Last edited: Sep 20, 2009
  20. Sep 21, 2009 #19
    Thanks! This forum is so awesome! It's nice to have a place to discuss these things. Thank you to everyone for your insights. Sylas, thanks for pointing out my errors.

    Does anyone know of a way to change the title of this thread? A more descriptive title may help others who are searching for solutions to similar problems.

    Thanks again guys, this one has been bugging me for awhile now.
     
  21. Sep 21, 2009 #20

    sylas

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    You're welcome. I don't think the title will matter much; I very much doubt that people look through old threads in search of similar problems.

    Felicitations -- Sylas
     
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