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anonymous

  • 5 years ago

how do you find the equation of a tangent line through a point outside a circle. Suppose the point is (6,-5) and the circle is (x-1)^2 + (y^2) = 4 , but i want to find a general formula. we know that two tangents can always be drawn to a circle from any point outside the circle

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  1. amistre64
    • 5 years ago
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    so like this?

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  2. anonymous
    • 5 years ago
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    YESSSSSSSSS!!!!!!!!!!!!!!

  3. anonymous
    • 5 years ago
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    in an xy plane

  4. amistre64
    • 5 years ago
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    since a tangent to the circle is always at 90 degrees to the radius.... this makes a right triangle.

  5. anonymous
    • 5 years ago
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    hey how do you draw attachments like that

  6. amistre64
    • 5 years ago
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    the hyp is the length of the center of the cirlce to the point and one leg is the radius

  7. amistre64
    • 5 years ago
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    i use paint in the windows accessories...save it to my desktop and retrieve it from there

  8. anonymous
    • 5 years ago
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    you can draw circles?

  9. amistre64
    • 5 years ago
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  10. amistre64
    • 5 years ago
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    yes; its the oval selection and hold down the shift key while you drag

  11. anonymous
    • 5 years ago
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    yes what program is it? windows paint?

  12. anonymous
    • 5 years ago
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    ok so we have a point, lets call it (x0,y0) outside a circle

  13. amistre64
    • 5 years ago
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    yes; windows generi paint program that comes in every installation of windows since 3.xx

  14. anonymous
    • 5 years ago
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    ok i will check that

  15. anonymous
    • 5 years ago
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    i want to actually find a general formula here

  16. anonymous
    • 5 years ago
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    ok so we have (x0,y0) outside a circle (x-h)^2 + (y-k)^2 = r^2

  17. amistre64
    • 5 years ago
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    the general formula is the pythag thrm... to find a vector

  18. amistre64
    • 5 years ago
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    find a vector that is parallel to the tangent from the point....

  19. anonymous
    • 5 years ago
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    you cant do this analytically

  20. amistre64
    • 5 years ago
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    you can ; but then convert it :)

  21. anonymous
    • 5 years ago
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    we have y - y0 = m(x-x0) ,

  22. amistre64
    • 5 years ago
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    the slope of the tangent to the circle at any given point is f'(x)

  23. anonymous
    • 5 years ago
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    also the line from the center of circle to (x,y) is y-k = -1/m (x-h)

  24. anonymous
    • 5 years ago
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    the normal i mean

  25. anonymous
    • 5 years ago
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    since (x,y) is unknown

  26. amistre64
    • 5 years ago
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    given a standard point on the circle for say (Xc,Yc)...

  27. anonymous
    • 5 years ago
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    oh then i messed up

  28. amistre64
    • 5 years ago
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    the point has to be within a certain domain for an outside point to have a tangent right?

  29. anonymous
    • 5 years ago
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    thats possible

  30. anonymous
    • 5 years ago
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    there are 3 points given

  31. anonymous
    • 5 years ago
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    (h,k) the center of the circle, (xo,yo) point outside circle (x1,y1) the point on the circle and (x2,y2) the second point on circle that intersects tangent

  32. anonymous
    • 5 years ago
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    so 4 points

  33. amistre64
    • 5 years ago
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    f'(c) that passes thru P(x1,y1) in the field and P1(xc,yc) on the circle

  34. amistre64
    • 5 years ago
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    if we view this from the origin...would it be simpler?

  35. amistre64
    • 5 years ago
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    the angle formed by the center to outside point forms an angle [a] with the x axis; the angle between that line and the radius to tangent form [b] right

  36. anonymous
    • 5 years ago
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    yes

  37. amistre64
    • 5 years ago
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    like this so far

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  38. amistre64
    • 5 years ago
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    a' is in the wrong spot lol

  39. anonymous
    • 5 years ago
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    so you can do this vectorially? or with derivatives?

  40. amistre64
    • 5 years ago
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    you can, but lets get a clear picture first and then add the details :)

  41. amistre64
    • 5 years ago
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    the perp lines slope we want is tan(a+b) right?

  42. anonymous
    • 5 years ago
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    tan a + b ?

  43. amistre64
    • 5 years ago
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    a = cos^-1(CP/r)

  44. amistre64
    • 5 years ago
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    yes; from my picture I got a + b = the slope of the line that is perp to the tangent....

  45. amistre64
    • 5 years ago
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    and tan(a+b) givs us that slope; so figure the angles... :)

  46. amistre64
    • 5 years ago
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    does this makes sense?

  47. anonymous
    • 5 years ago
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    no

  48. amistre64
    • 5 years ago
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    we center the circle at the origin so we can examine it...

  49. anonymous
    • 5 years ago
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    ok

  50. anonymous
    • 5 years ago
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    i dont understand the a = arcos (CP/r) business, where does that come from

  51. amistre64
    • 5 years ago
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    the angle made by the line from the center to the extPoint makes an angle [a] with the x axis...that angle is cos^-1[]

  52. anonymous
    • 5 years ago
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    ok , the angle is ?

  53. amistre64
    • 5 years ago
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    the angle is cos(a) = P.r/(|P||r|)...dot product of 2 vecotrs

  54. anonymous
    • 5 years ago
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    cos a = Ca' / ( CP)

  55. amistre64
    • 5 years ago
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    P being the outside point

  56. anonymous
    • 5 years ago
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    what is r ?

  57. amistre64
    • 5 years ago
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    cos(b) = T.P/(|P||r|)

  58. amistre64
    • 5 years ago
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    radius = r.....

  59. anonymous
    • 5 years ago
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    cos b =r / TP

  60. amistre64
    • 5 years ago
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    to find the tangent..... we use the pythag thrm; |T|=sqrt(|P|^2-r^2)

  61. anonymous
    • 5 years ago
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    ok but we want the tangent line equation

  62. anonymous
    • 5 years ago
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    |T| is infinity

  63. anonymous
    • 5 years ago
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    but that line segment, right comes from pythangrean

  64. anonymous
    • 5 years ago
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    and you were doing the inner product, im a bit rusty

  65. amistre64
    • 5 years ago
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    back...

  66. anonymous
    • 5 years ago
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    hey

  67. amistre64
    • 5 years ago
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    so if we know tha toutside point ot not; we can assign a vector to it ..p <x,y> we can also assign a vector to our radius at any given point [c]... r = <c,f(c)>. when f'(c) is parallel with r-p..... we have our tangent line

  68. amistre64
    • 5 years ago
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    the equation of a tangent line thru any point is given by: x = x0 + at y = y0 + bt z = 0....since we are in the xy plane

  69. anonymous
    • 5 years ago
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    oh i think i have it

  70. anonymous
    • 5 years ago
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    if youre given a circle and a point outside a circle

  71. anonymous
    • 5 years ago
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    nevermind

  72. amistre64
    • 5 years ago
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    ...... but i gotta mind lol

  73. amistre64
    • 5 years ago
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    if we are examine it from the origin; like all good analysts; we move the point and the circle by -x and -y..... make sure to write that in a note so we know were to put it when were done...

  74. anonymous
    • 5 years ago
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    i guess im confused on a lot of things. where did you get tan (a+b)

  75. anonymous
    • 5 years ago
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    where did you get arcos (cP/r)

  76. amistre64
    • 5 years ago
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    i was rummaging thru the thoughts in my head.... some were good, others not so good :)

  77. anonymous
    • 5 years ago
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    where did point c come from ?

  78. anonymous
    • 5 years ago
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    ok youre starting over

  79. amistre64
    • 5 years ago
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    c = center; ..... or origin depending on where the circle was

  80. amistre64
    • 5 years ago
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    since we dont know the point ot tangency, i figure vectors are a safe way to find it

  81. anonymous
    • 5 years ago
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    yeah thats the trouble

  82. amistre64
    • 5 years ago
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    we know the origin; we have vectors splayed out from there. the vector for the radius is simply <c,f(c)> where c is the point of tangency right?

  83. amistre64
    • 5 years ago
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    p is the vector from the origin to the outside point..<x,y>

  84. amistre64
    • 5 years ago
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    we know that if we subtract p from r we get a new vector heading inthe direction of tangency...

  85. amistre64
    • 5 years ago
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    it has to match f'(c) in order to determine if 'c' is a match..

  86. anonymous
    • 5 years ago
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    one sec

  87. amistre64
    • 5 years ago
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    when we find the right [c] we have the f'(c) as the slope for our line; and we have a point to place in it to figure out the b of y=f'(c)x + b

  88. anonymous
    • 5 years ago
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    ok vector R - P has the same direction as the tangent line , i agree

  89. anonymous
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    because P + (R-P) = R , where R is that point (c,f(c)) or the vector to the point

  90. anonymous
    • 5 years ago
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    so f ' (c) will , you mean the vector c ? or just

  91. amistre64
    • 5 years ago
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    the derivative of the function itself.

  92. anonymous
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    so f ' ( < c , f(c) ) >

  93. anonymous
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    oh , the derivative of the equation of the circle

  94. amistre64
    • 5 years ago
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    the circle is what i had in mind; but the vector might work just as good

  95. anonymous
    • 5 years ago
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    2x + 2y * y' = 0

  96. anonymous
    • 5 years ago
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    i prefer analytic, my vector skills are shaky :)

  97. anonymous
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    so y ' = -x/y

  98. amistre64
    • 5 years ago
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    vectors are analytic; just useful when drawn as arrows lol

  99. anonymous
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    true

  100. anonymous
    • 5 years ago
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    ok i prefer baby algebra ?

  101. amistre64
    • 5 years ago
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    lets restrict our selves to half the circle..... for simplicity y = sqrt(r^2 - x^2)

  102. anonymous
    • 5 years ago
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    its confusing, because the points , like an n tuple, can be represented as a geometric point, or an arrow from the origin,

  103. anonymous
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    ok

  104. anonymous
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    so f ' (c) = -c / ( sqrt ( r^2 - c^2 )

  105. anonymous
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    you must be good at geometry, im better with algebra/symbols, that sort of thing

  106. anonymous
    • 5 years ago
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    so we have x = tv + x0 or something

  107. anonymous
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    x = x0 + at , y = y0 + at , but then we will have to remove the t

  108. amistre64
    • 5 years ago
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    the derivative of u^(1/2) = 1/[2u^(1/2)] u = 1-x^2 ; du = -2x the derivative of sqrt(r^2 - x^2)...so i agree with the derivative :)

  109. amistre64
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    the t is just a scalar..

  110. amistre64
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    it can be 1 for all it cares

  111. anonymous
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    well i want the final solution to look like y = mx + b

  112. anonymous
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    the problem is that (c,f(c) is unknown

  113. amistre64
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    c has to meet the criteria set forth.... once it meets it; we know it :)

  114. radar
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    will it meet it today?

  115. amistre64
    • 5 years ago
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    lol...depends on how it behaves lol

  116. amistre64
    • 5 years ago
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    if anything ; f'(c) is the slope of the line passing thru points (c,f(c)) and (x,y)

  117. anonymous
    • 5 years ago
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    ok

  118. amistre64
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    y = f'(c)x + B

  119. anonymous
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    ok

  120. amistre64
    • 5 years ago
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    ive been playing some numbers here... see if this follows... p =<6,5> ; r = 2 R=<c,f(c)> f'(c) = -1/sqrt(2^2 - c^2) R-p = <c-6,2-5> = (c-6,-3)

  121. amistre64
    • 5 years ago
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    -c -3 --------- = ---- sqrt(4-c^2) c-6

  122. amistre64
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    cross multiply... -c(c-6) = -3(sqrt(4-c^2)) ; /-3 -c^2 +6c -------- = sqrt(4-c^2 ; ^2 both sides.. -3

  123. amistre64
    • 5 years ago
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    c^4 -12c^3 +36c^2 ------------------ = 4-c^2 9

  124. amistre64
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    *9 both sides c^4 -12c^3 +36c^2 = 36 - 9c^2

  125. amistre64
    • 5 years ago
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    zero out the rhs... c^4 -12c^3 +9c^2 + 36c^2 -36 = 0 c^4 -12c^3 + 45c^2 -36 = 0

  126. anonymous
    • 5 years ago
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    how did you get this -c -3 -3 --------- =----------- sqrt(4-c^2) c-6

  127. anonymous
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    there are some algebraic approaches on yahoo answers, but im not sure they are correct

  128. anonymous
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    http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba

  129. amistre64
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    the derivative at c = -c/sqrt(r^2-c^2) right?

  130. amistre64
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    at r = 2....

  131. amistre64
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    ...... i see a hiccup tho; i used (c,2) instead of (c,f(c)).... gonna have to rewrite it al lol

  132. amistre64
    • 5 years ago
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    -c sqrt(4-c^2)-5 ---------- = ------------- sqrt(4-c^2) c-6

  133. amistre64
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    -c(c-6) = sqrt(4-c^2)[sqrt(4-c^2)-5] -c^2 +6c = (4-c^2) -5sqrt(4-c^2) 6c -4 ------ = sqrt(4-c^2) -5

  134. amistre64
    • 5 years ago
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    now ^2 both sides :) 6c^2 -48c +16 = 25(4-c^2) 6c^2 -48c +16 = 100 -25c^2

  135. amistre64
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    36c^2....

  136. amistre64
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    36c^2 +25c^2 -48c +16 -100 = 0 61c^2 -48c -84 = 0

  137. amistre64
    • 5 years ago
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    solve for c

  138. anonymous
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    hmmm

  139. anonymous
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    well this does seem easier than the yahoo method , its insanely difficult

  140. anonymous
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    if you want to read it its here http://answers.yahoo.com/question/index?qid=20090721083008AA1VEba

  141. anonymous
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    YOU GET MEDAL for persistence !!!

  142. amistre64
    • 5 years ago
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    :) i get from teh wolfram graphing ; that c = -.844 and 1.631

  143. anonymous
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    one sec, let me draw this

  144. anonymous
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    wait

  145. anonymous
    • 5 years ago
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    so you said the direction of R- P where R-P = <a,b> say is b/a >

  146. anonymous
    • 5 years ago
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    i think theres something wrong with that , oh, it only works in 2 dimensions

  147. amistre64
    • 5 years ago
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    lets define p as (a,b) thats fine... but we dont want to confuse it for the quadratic.... Ac^2 +Bc + C (a^2 -b^2) = A (-2ra) = B [r^2(1+b^2)] = C

  148. amistre64
    • 5 years ago
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    ...got some signs wrong lol

  149. anonymous
    • 5 years ago
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    so basically you did f ' (c) = slope [ segment RP ]

  150. anonymous
    • 5 years ago
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    and we know that f ' (c) from the implicit derivative -x/y

  151. amistre64
    • 5 years ago
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    yes

  152. anonymous
    • 5 years ago
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    oh well then you didnt really need to use vectors then , i guess

  153. amistre64
    • 5 years ago
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    I used vectors to compare it inititally to get a formula for the rest... (a^2 -b^2) = A (-2ra) = B [r^2(1+b^2)] = C

  154. anonymous
    • 5 years ago
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    in higher dimensions we will need vectors

  155. amistre64
    • 5 years ago
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    vector <c-a, f(c)-b> is used to calibrate the slope

  156. anonymous
    • 5 years ago
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    but you took the slope of that vector? correct ?

  157. amistre64
    • 5 years ago
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    (a^2 +b^2) = A (-2ra) = B [r^2(1-b^2)] = C

  158. amistre64
    • 5 years ago
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    correct, the vector x and y form a slope.... so yes

  159. amistre64
    • 5 years ago
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    c = thats stuf in the quadraitc formula lol

  160. anonymous
    • 5 years ago
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    whats all this Ac^2 +Bc + C (a^2 -b^2) = A (-2ra) = B [r^2(1+b^2)] = C

  161. amistre64
    • 5 years ago
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    i took the numbers out that i was using and placed the (a,b) fromthe exteroir point back into it...

  162. amistre64
    • 5 years ago
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    the point of tangency is at c..which is equal to the solution to the quadratic equation formed by .... (a^2+b^2)c^2 -(2ra)c + [r^2(1-b^2)] = 0

  163. anonymous
    • 5 years ago
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    youre making a general solution

  164. amistre64
    • 5 years ago
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    yes :)...its no good if it only used for one point

  165. anonymous
    • 5 years ago
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    ok start over, where did you A, B< and C

  166. amistre64
    • 5 years ago
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    ..... when i used the the point (6,5) and a radius of 2... i manipulated the algebra to get a final quadratic form right?

  167. amistre64
    • 5 years ago
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    61c^2 -48c -84 = 0 [c] can be solved by teh quadratic formula now since we know a b and c

  168. amistre64
    • 5 years ago
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    61 is from a^2 + b^2; 36 +26 = 61

  169. amistre64
    • 5 years ago
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    -48 is fron -2ra; -2(2)(6).....gonna have to review tha tone lol and -84 is from (r^2[1-b^2]): 4(1-25) = 4-100......murmur..

  170. anonymous
    • 5 years ago
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    ok

  171. amistre64
    • 5 years ago
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    paper is so much easier to work on.... Ac^2 +Bc +C [a^2+b^2] = A [-2a(r^2)] = B [r^2(r^2-b^2)] = C

  172. amistre64
    • 5 years ago
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    (a,b) (6,5) ; r=2 (36+25) = 61 = A -2(6)(2^2) = -12(4) = -48 = B 4(4-25) = 16 - 100 = -84 = C

  173. anonymous
    • 5 years ago
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    ok starting with a point (a,b)

  174. anonymous
    • 5 years ago
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    external to the circle, and the point tangent on the circle is (c,f(c)) and the radius is r , center , well we used (0,0)

  175. amistre64
    • 5 years ago
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    to find the point [c] of tangency we get: c = : ar^2 sqrt[4a^2r^4 - 4(a^2+b^2)(r^2(r^2-b^2))] ------- ( +-) ----------------------------- a^2+b^2 a^2+b^2

  176. amistre64
    • 5 years ago
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    if the center is not at the origin; then use transformations to get it there :)

  177. anonymous
    • 5 years ago
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    actually this wont work

  178. anonymous
    • 5 years ago
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    -c / sqrt ( r^2 -c^2) = ...

  179. anonymous
    • 5 years ago
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    the right side could be b - f(c) or f(c) - b

  180. anonymous
    • 5 years ago
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    we used (f(c) - b ) / ( c- a)

  181. anonymous
    • 5 years ago
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    so you can say, if f(c) > b, then ...

  182. amistre64
    • 5 years ago
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    its the same slope regardless of the way its pointing

  183. anonymous
    • 5 years ago
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    oh right , duh

  184. anonymous
    • 5 years ago
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    so then we get -c / sqrt ( r^2 -c^2) =(f(c) - b ) / ( c- a)

  185. anonymous
    • 5 years ago
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    -c / sqrt ( r^2 -c^2) =(srt (r^2 - c^2) - b ) / ( c- a)

  186. amistre64
    • 5 years ago
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    the whole point is to get to a point c to find the derivative to use in the equation to the tangent line lol

  187. anonymous
    • 5 years ago
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    correct

  188. anonymous
    • 5 years ago
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    but i get a simpler derivation

  189. amistre64
    • 5 years ago
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    but we have the slope; and we have a point from the vectors...

  190. anonymous
    • 5 years ago
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    -c / sqrt ( r^2 -c^2) =(srt (r^2 - c^2) - b ) / ( c- a) multiply both sides by sqrt ( r^2 - c^2)

  191. anonymous
    • 5 years ago
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    -c (c-a) =(r^2 - c^2) - b sqrt (r^2-c^2)

  192. anonymous
    • 5 years ago
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    actually i cross multiplied , ok so far?

  193. amistre64
    • 5 years ago
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    good so far

  194. anonymous
    • 5 years ago
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    sqrt (r^2 - c^2) = r^2/b - ac/b

  195. amistre64
    • 5 years ago
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    i think we broke openstudy... the fonts gotten weird on my end

  196. anonymous
    • 5 years ago
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    you can start over, use www.openstudy.com , and click math

  197. anonymous
    • 5 years ago
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    r^2 - c ^2 = ( r^2 - ac)^2 / b^2

  198. anonymous
    • 5 years ago
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    -c (c-a) =(r^2 - c^2) - b sqrt (r^2-c^2) sqrt (r^2 - c^2) =( r^2 - ac)/b ok so far?

  199. anonymous
    • 5 years ago
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    oh i subtracted c^2 from both sides by the way

  200. amistre64
    • 5 years ago
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    so did i :)

  201. anonymous
    • 5 years ago
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    ok squaring both sides i get r^2*b^2 - b^2 c^2 = r^4 -2r^2ac + a^2 *c^2

  202. anonymous
    • 5 years ago
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    now r is given, b and a are given

  203. anonymous
    • 5 years ago
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    i didnt get A = a^2 +b ^2 though

  204. anonymous
    • 5 years ago
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    still there?

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