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Line, Circle, Locus problem

  1. Apr 27, 2015 #1
    1. The problem statement, all variables and given/known data
    If the line ## hk + ky = 1 ## touches the circle ## x^2 + y^2 = a^2 ##, then locus of the points (h,k) is a circle of radius :
    a
    h
    k
    1/a

    2. Relevant equations
    NA
    3. The attempt at a solution
    Tried differentiating circle equation (Is differentiation of use here?, also don't know if it is a calculus or pre-calculus problem. )
    Got y' = -x/y
    Differentiated line equation and thinking h,k are constants which is not true I got y' = 0
    I have to substitute something but what?
     
    Last edited: Apr 27, 2015
  2. jcsd
  3. Apr 27, 2015 #2

    BvU

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    Make a drawing !
    You know the slope of the line, you know an equation for a point it has to go through.

    And you probably have a typo in the equation for the line ?
     
  4. Apr 27, 2015 #3
    No, I have not a typo in equation of line.
    The question is confusing me as I think there should be x in equation of line?
     
  5. Apr 27, 2015 #4

    BvU

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    Now it says ##y={1\over k}-h## having slope zero. Touches the circle if ...
     
  6. Apr 27, 2015 #5
    Not getting, are you saying I have a typo in Question statement of line equation?
     
  7. Apr 27, 2015 #6

    BvU

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    That is what I said and that is what I typed. ##
    hk + ky = 1## is the equation for a horizontal line. Slope 0. Does it say that in the exercise you were given ?
     
  8. Apr 27, 2015 #7
    Oh, so you are seeing my attempt?
     
  9. Apr 27, 2015 #8

    BvU

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    Can we please agree that the exercise text states ##hk + ky = 1## as the equation of the line ? If so, we can proceeed and prove that the locus of the points (h,k) is not a circle.
     
    Last edited: Apr 27, 2015
  10. Apr 27, 2015 #9
    Yeah we can agree.
    Now how to proceed?
     
  11. Apr 27, 2015 #10

    BvU

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  12. Apr 27, 2015 #11

    HallsofIvy

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    Is it hk+ ky= 1 or hx+ ky= 1?
     
  13. Apr 27, 2015 #12

    BvU

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    In post #9 Raghav repeats hk+ ky= 1 and now we are under way to discover that that does NOT give a circle as locus. Read !
     
  14. Apr 27, 2015 #13
    y = a or y = -a.
    So the question is wrong as locus is not a circle but mere two points.
     
  15. Apr 27, 2015 #14

    BvU

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    Not the way it works. You are asked for the locus of the points (h,k), not for the points (x,y) where the line touches the circles.

    To make a long story short: ##y=\pm a## is correct as the equation for the tangent line. If you want to write that as ##hk + ky = 1## you get e.g. ##hk + ka = 1## so the points (h,k) satisfy ##k = {1\over h+a}## which is not the equation of a circle.

    Having delivered proof that ##hk + ky = 1## is not what was intended for this exercise, I propose you now accept that it should have been ##hx + ky = 1##.

    What do we know about these lines ?

    Well, as you already found in post #1, they have slope -x/y . Also, one of the points on the line satisfies ##x^2+y^2 = a^2##. So what do we have to impose on h and k to make the equation for such a line take the form ##hx + ky = 1## ?

    (My advice: draw a circle and a tangent :rolleyes: !)​
     
    Last edited: Apr 27, 2015
  16. Apr 28, 2015 #15
    Took a point on line x1,y1
    Then slope is - x1/y1.
    x12 + y12 = a2
    Also hx1 + ky1 = 1.
    Here is a circle of radius a and some random 2 tangents drawn.

    image.jpg
     
  17. Apr 28, 2015 #16

    BvU

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    Good start. The point where the line is tangent to the circle is (x,1y1).
    Now (important step): what's the equation of the line ? (I don't mean hx1 + ky1 = 1 but the equation for the line in therms of x1, y1 and a)
     
  18. Apr 28, 2015 #17
    I was doing some manipulations,
    Take the general equation

    ## hx + ky = 1 ##

    ## ⇒ y = \frac{1-hx}{k} ##

    In equation ## x^2 + y^2 = 1 ##

    Substituting y value as it satisfies,

    ## x^2 + \frac{1+ h^2x^2 -2hx}{k^2} = a^2 ##
    Rearranging,

    ## x^2( k^2 + h^2) - 2hx +1 - k^2a^2 = 0 ##

    It striked to me that

    Now discriminant of quadratic equation should be zero as tangent line touches one point.

    ## 4h^2 - 4 (k^2 + h^2)(1-k^2a^2) = 0 ##

    Solving we get,

    ## k^2 + h^2 = 1/a^2 ##
    So locus is circle of radius 1/a

    What was the use of diagram here and calculating slope earlier as -x/y ?
     
  19. Apr 28, 2015 #18

    BvU

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    LIne segment from origin to (x1, y1) has length a.
    Tangent line in (x1, y1) is perpendicular to that line.

    That means all points (x,y) on that tangent line satisfy $$ (x_1,y_1)\cdot(x, y) = a^2$$ -- an inner product, written out : $$x_1 x + y_1 y = a^2$$ -- and there's your tangent line equation without any quadratic equation at all !
    Compare $$x_1 x + y_1 y = a^2 \quad {\rm to} \quad x + k y = 1$$ and you see
    $$h = x_1/a^2 \quad {\rm and} \quad k = y_1/a^2$$
    Since $$ x_1^2+ y_1^2 =a^2$$ you indeed find
    $$ h^2+ k^2 =1/a^2$$

    Well done !

    I hope you are now also convinced that the line equation in post #1 was a typo :wink:?

    --
     
  20. Apr 28, 2015 #19
    Oh, another and easy way
    Not getting this
    $$ (x_1,y_1)\cdot(x, y) = a^2$$ -- an inner product,
     
  21. Apr 28, 2015 #20

    BvU

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    Have you ever heard of the normal equation for a straight line ? We learned it at school (I was 15 then...).
    It goes like this:

    Locus.jpg
    For all points on the dashed tangent: Inner product $$(x,y) \cdot\ (x_1, y_1) = \vec v_1 \cdot \vec v_2 = |\vec v_1 |\;|\vec v_2| \;\cos\phi = |\vec v_1 |\;|\vec v_1|/ \cos\phi \;\cos\phi = |\vec v_1 |^2$$
    Note that this time the arrows aren't for infinite lines but for vectors.

    There is also a unit normal vector form: if ##\tan \alpha = y_1/x_1## (hey, that's like (-x_1/y_1) -1 !)

    you have with ##\vec n = (x_1, y_1), \quad \hat n = (\cos\alpha, \sin\alpha) = (x_1, y_1) / |\vec v_1| ## that $$ x\cos\alpha + y\sin\alpha = |n| \quad {\rm or} \quad \vec v \cdot \hat n = |\vec n|$$
    --
     
    Last edited: Apr 28, 2015
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