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anonymous

  • one year ago

How do you find the equation of a tangent line and also normal line when given an equation of the circle and a point that is NOT on the circle .. the problem x^2+y^2=13 and point (-4, 7) "find the equation of a tangent line and also normal line when given an equation of the circle and a point that is NOT on the circle"

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  1. MrNood
    • one year ago
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    |dw:1444258760551:dw| The line that is normal is always a diameter, and passes through the centre. From the equation of th ecircle you know the cntre is (0,0) so you have 2 points on the normal line so can work out its equation. The slope of the tangent is perpendicular to the normal so its slope is -1/m

  2. MrNood
    • one year ago
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    because the point is inside the circle there is no tangent through the point...

  3. anonymous
    • one year ago
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    no the point is outside the circle ...this is how it look..

  4. anonymous
    • one year ago
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    1 sec

  5. MrNood
    • one year ago
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    ok - sorry - i see my mistake:

  6. MrNood
    • one year ago
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    |dw:1444259449121:dw|

  7. anonymous
    • one year ago
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  8. anonymous
    • one year ago
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    yea the lecture give us it and told us to finish it off... stww.. idk what to do..

  9. anonymous
    • one year ago
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    @amistre64

  10. amistre64
    • one year ago
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    do you know how to determine the equation of a tangent line using derivatives?

  11. amistre64
    • one year ago
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    or, we can leave the slope open and solve it after equating the line and the circle equations

  12. anonymous
    • one year ago
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    no .. this is geometry ...but if its easier y not show mee

  13. amistre64
    • one year ago
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    what would the equation for a line going thru a point having a slope .. be?

  14. anonymous
    • one year ago
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    y=mx+b ??

  15. amistre64
    • one year ago
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    thats only if we have a special point, the intercept. in general we can use the point slope format y-b = m(x-a) or simply y = m(x-a)+b for a point (a,b) now lets sub this into our circle equation where y is

  16. amistre64
    • one year ago
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    x^2+y^2=13 x^2+(m(x-a)+b)^2=13 expand it out

  17. amistre64
    • one year ago
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    you can use -4,7 instead of a,b of course

  18. anonymous
    • one year ago
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    ok x^2+(m(x+4)+7)^2=13

  19. anonymous
    • one year ago
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    ok x^2+(m(x+4)+7)^2=13

  20. amistre64
    • one year ago
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    now expand it all out ... im working it on notepad to see if we come to the same conclusions :)

  21. anonymous
    • one year ago
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    do i find whats m first?

  22. anonymous
    • one year ago
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    m=7/-4

  23. amistre64
    • one year ago
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    m is undetermined, it is something we will hopefully find out at the end of the process.

  24. anonymous
    • one year ago
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    o ok

  25. anonymous
    • one year ago
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    x^2+(m(x+4)+7)^2=13 x^2+(mx+4m+7)^2=13 x^2+m^2x^2+16m^2+49=13

  26. anonymous
    • one year ago
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    x^2(1+m^2)+16m^2=-36

  27. amistre64
    • one year ago
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    x^2 + (m(x+4)+7)^2 = 13 x^2 + m^2(x+4)^2 +49 +14m(x+4) = 13 x^2 + m^2(x^2+16+8x) +49 +14mx +56 = 13 x^2 + m^2 x^2 +16m^2 +8m^2x +92 +14mx = 0 (m^2+1) x^2 +(8m^2+14m) x +(9 +16m^2) = 0 in order for there to only be one root, the discriminant of this quadratic need to equal zero. (8m^2+14m)^2 -4(m^2+1)(9 +16m^2) = 0 64m^4 +196 m^2 +2(8)(14)m^3 -(4m^2+4)(9 +16m^2) = 0 64m^4 +196 m^2 +2(8)(14)m^3 -(36m^2 +36 +64m^4 +64m^2) = 0 64m^4 +196 m^2 +2(8)(14)m^3 -36m^2 -36 -64m^4 -64m^2 = 0 2(8)(14)m^3 +96 m^2 -36 = 0 i cant be certain that i kept all of that straight :)

  28. anonymous
    • one year ago
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    x^2 + m^2(x+4)^2 +49 +14m(x+4) = 13 where did u get the 14m(x+4) part frm?

  29. amistre64
    • one year ago
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    gotta rechk that, the wolf says im off i think ... (a+b)^2 = a^2 + b^2 + 2ab if a=m(x+4) and b=7

  30. amistre64
    • one year ago
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    and (nm)^2 = n^2 m^2 so im wondering why the wolf says m(x+4)^2 instead of m^2(x+4)^2

  31. anonymous
    • one year ago
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    which wolf are u talking about...

  32. amistre64
    • one year ago
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    wolframalpha.com

  33. amistre64
    • one year ago
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    mx + 4m + 7 mx + 4m + 7 ------------ mmxx +4mmx +7mx 4mmx +16mm + 28m 7mx + 28m +49 ------------------------------------- m^2x^2 +8m^2x +14mx +16m^2 +56m + 49 m^2(x^2 +8x +16) + m(14x+56) + 49

  34. anonymous
    • one year ago
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    shouldnt it be mmxx+ 8mm+14 im confuse ...were u adding it?

  35. amistre64
    • one year ago
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    that is NOT how you multiply

  36. amistre64
    • one year ago
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    i should prolly ask how you ahve been solving these before i came in to muddy the waters

  37. anonymous
    • one year ago
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    (mx+4m+7)(mx+4m+7) mxmx+mx4m+7mx+4mmx+16mm+28m+7mx+28m+49 m^2x^2+4m^2x+7mx+4m^2x+16m^2+28m+7mx+28m+49 m^2x^2+8m^2x+14mx+16m^2+56m+49

  38. amistre64
    • one year ago
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    |dw:1444263458406:dw| what is the value of n?

  39. amistre64
    • one year ago
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    n^2 = 4^2+7^2-13 = 52 n = sqrt(52) right?

  40. amistre64
    • one year ago
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    now we know the radius of a circle centered at -4,7 and can equate 2 circles to find where they meet (x+4)^2 + (y-7)^2 - 52 = x^2 + y^2 -13

  41. anonymous
    • one year ago
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    ok i will be back in the next 20 second ...

  42. amistre64
    • one year ago
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    i got it figured out :) ill be back as well ... food is on the table

  43. anonymous
    • one year ago
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    but how can the circle be center at (-4,7) when thats the point the tangent pass through and that point is not on the circle ...this was what the lecture give... and then tell us figure our the rest...

  44. amistre64
    • one year ago
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    you can center a circle at any point you wish ... my solution is to create a circle at the given point, with a radius that reaches to the point of tangency ... then solve the 2 equations to see where their solution set is

  45. amistre64
    • one year ago
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    |dw:1444264864450:dw|

  46. anonymous
    • one year ago
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    o ok looks interesting

  47. amistre64
    • one year ago
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    so, i would determine the distance from the given point to the origin (the center of our given circle) as it forms the hypotenuse of a right triangle with the radius of the given circle as a leg. the radius of our unknown circle is the missing leg |dw:1444264994717:dw|

  48. amistre64
    • one year ago
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    n = sqrt(52) so we our circle of radius sqrt(52) centered at -4,7 is (x+4)^2 + (y-7)^2 = 52 and equate this with the given: x^2 + y^2 = 13 (x+4)^2 + (y-7)^2 +13 = 52 +x^2 + y^2 x^2+8x+16 + y^2-14y+49 +13 = 52 +x^2 + y^2 8x+16 -14y+49 +13 = 52 we have reduced the solution to a linear equation; solve for y 8x -14y= 52 -16 -13 -49 8x -14y= -26 4x -7y= -13 i wonder of that is a coincidence ... y = (4x+13)/7

  49. amistre64
    • one year ago
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    this is actually the line the passes between the points of tangency ... it makes sense since 2 points create a line to start with and the solution set is 2 points. sub this in to find x x^2 + y^2 = 13 x^2 + [(4x+13)/7]^2 = 13

  50. amistre64
    • one year ago
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    |dw:1444265497662:dw|

  51. anonymous
    • one year ago
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    im not sure if i follow all this ..but x^2+(4x+13)^2/49=13 49x^2+16x^2+104x+169=637 65x^2+104x-468=0 5x^2+8x-13=0 5x^2-5x+13x-13=0 5x(x-1)+13(x-1)=0 5x+13=0 x=-13/5 and x-1=0 x=1

  52. amistre64
    • one year ago
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    d^2 = (a^2+b^2), for some point (a,b) n^2 = a^2+b^2 - k, for some given r^2=k (x-a)^2 + (y-b)^2 +k = n^2 + x^2 + y^2 x^2 -2ax + a^2 + y^2 -2by +y^2 +k = n^2 + x^2 + y^2 -2ax + a^2 -2by +b^2 +k = n^2 -2ax + a^2 +b^2 +k - n^2 = 2by [-2ax +a^2 +b^2 +k -(a^2+b^2 - k)]/2b = y [-2ax +2k]/2b = y [-ax +k]/b = y

  53. amistre64
    • one year ago
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    x=-18/5 and x=2 i think you might have made a few errors

  54. amistre64
    • one year ago
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    so it does generalize into a line solution of y = (-ax+k)/b y = (4x+13)/7 in this case

  55. amistre64
    • one year ago
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    x^2 + (4x+13)^2/49 = 13 49x^2 + 16x^2 +169 +8(13)x = 13(49) 65x^2 +104x -468 = 0 13 (x-2) (5x+18) = 0

  56. amistre64
    • one year ago
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    using y = (4x+13)/7 with those x values we can solve for y ... and all this does is gives us the tangency points which we can then find the slopes (m1,m2) for the lines y = m1(x+4) + 7 and y = m2(x+4) + 7

  57. amistre64
    • one year ago
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    or as noted, y' = -x1/y1 will give us the slopes as well

  58. anonymous
    • one year ago
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    omg all this working to 1 question...i wonder how much marks if this question come as a test question in the next 2 weeks..would it go for...stw... hope this dont come...any how what does m1=? did we find that?

  59. amistre64
    • one year ago
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    you have to determine your points of tangency ... we found the x values (-18/5 and 2), what are the y values ?

  60. anonymous
    • one year ago
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    y=m1(x+4)+7 when x=-18/5 y=m1(-18/5+4)+7 y=m1(2/5)+7 am i going right?

  61. amistre64
    • one year ago
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    no, the solution set is on the line: y = (-ax+k)/b y = (4x+13)/7 when x=2 ... y = 21/7 = 3 when x=-18/5 ... y = (4(-18)/5+13)/7 = ??

  62. amistre64
    • one year ago
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    the slope of that tangent at 2,3 is: -2/3 , the normal slope is 3/2

  63. anonymous
    • one year ago
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    i got y=-1/5

  64. amistre64
    • one year ago
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    correct

  65. amistre64
    • one year ago
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    y = m(x-a)+b x^2 + m^2 (x^2 -2ax + a^2) + b^2 +2mb(x-a) = k x^2 + m^2 x^2 -2a m^2 x + (am)^2 + b^2 +2mb x- 2mba -k = 0 (m^2+1) x^2 +2m(b-am) x + (am)^2 + b^2 - 2mba -k = 0 this has one root value if the discriminant is equal to 0 soo (2m(b-am))^2 -4((m^2+1))((am)^2 + b^2 - 2mba -k) = 0 when we let a=-4, and b=7 (2m(7-(-4)m))^2 -4((m^2+1))(((-4)m)^2 + 7^2 - 2(7)(-4)m -13) = 0 the wolf simplifies that to m=-2/3 and m=-18 for our slopes

  66. anonymous
    • one year ago
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    ok... 1 sec brb

  67. amistre64
    • one year ago
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    (m(x-a)+b)^2 x^2 + m^2 (x^2 -2ax + a^2) + b^2 +2mb(x-a) = k x^2 + m^2 x^2 -2a m^2 x + (am)^2 + b^2 +2mb x- 2mba -k = 0 (m^2+1) x^2 +2m(b-am) x + (am)^2 + b^2 - 2mba -k = 0 (2m(b-am))^2 -4((m^2+1))((am)^2 + b^2 - 2mba -k) = 0 (2m(7-(-4)m))^2 -4((m^2+1))(((-4)m)^2 + 7^2 - 2(7)(-4)m -13) = 0 4m^2 (b-am)^2 -(4m^2+4)((am)^2 + b^2 - 2mba -k) = 0 (am)^2 + b^2 - 2mba -k 4m^2 + 4 ------------------------ 4a^2m^4 +4ab^2 m^2 -8abm^3 -4km^2 +4a^2 m^2 +4ab^2 -8abm -4k 4(a^2) m^4 -4(2ab) m^3 +4(a^2+ab^2-k) m^2 -4(2ab) m +4(ab^2-k) divide off the 4 (a^2) m^4 -(2ab) m^3 +(a^2+ab^2-k) m^2 -(2ab) m +(ab^2-k) ---------------- equals 0 yeah, i dont think there is anyway that you would want to work the formula for finding a roots of a quartic equation.

  68. amistre64
    • one year ago
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    might be some calc errors in there cause the wolf says we obtain a quadratic with your a,b

  69. amistre64
    • one year ago
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    oh, thats just the 4ac part ... negate and add in the b^2 might help

  70. amistre64
    • one year ago
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    4m^2 (b-am)^2 4m^2 (b^2 +(am)^2 -2abm) 4(a^2) m^4 -4(a^2) m^4 -4(2ab) m^3 +4(2ab) m^3 4 (b^2) m^2 -4(a^2+ab^2-k) m^2 +4(2ab) m -4(ab^2-k) -(a^2+ab^2-b^2-k) m^2 +(2ab) m -(ab^2-k) \[m=\frac{ab\pm\sqrt{a^2b^2-(a^2+ab^2-b^2-k)(ab^2-k)}}{a^2+ab^2-b^2-k}\] not too sure how useful this would be ... doesnt look simple to memorize :) \[m=\frac{7(-4)\pm\sqrt{16(49)-(16-4(49)-49-13)(-4(49)-13)}}{16-4(49)-49-13}\]

  71. amistre64
    • one year ago
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    id stick with that: y = (-ax+k)/b and plugging that into the circle

  72. anonymous
    • one year ago
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    ok y do we have to do this part ...what are we finding for ? frm my last post to this post...

  73. amistre64
    • one year ago
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    we are finding the x and y values for the points of tangency to the given circle so that we can define sloped for the lines of tangency/normality

  74. amistre64
    • one year ago
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    if you know of some other way ... feel free

  75. anonymous
    • one year ago
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    no urs is ok... ill figure it out... so whats the next step... and i really appreciate ur help

  76. amistre64
    • one year ago
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    thru generalization, i determined that the solution points will form the line: y = (-ax+k)/b this is the line that connects the 2 points of tangency; does this make sense?

  77. amistre64
    • one year ago
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    |dw:1444270137860:dw|

  78. anonymous
    • one year ago
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    ok like this |dw:1444266705960:dw|

  79. amistre64
    • one year ago
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    yes, those are the line equations we need to determine, but we need to find the points of tangency on the circle using the point -4,7 and a circle equal to 13 y = (4x+13)/7 will help us get our points of tangency x^2 + y^2 = 13, sub in for y x^2 + [(4x+13)/7]^2 = 13 we determined that x=2 and -18/5 y = (4(2)+13)/7 = 21/7 = 3; (2,3) is one point y = (4(-18/5)+13)/7 = -1/5; (-18/5, -1/5) is the other point

  80. amistre64
    • one year ago
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    good to there?

  81. amistre64
    • one year ago
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    works coming early in the morning so i have to wrap this up soon

  82. anonymous
    • one year ago
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    yea good up to there

  83. amistre64
    • one year ago
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    since we know our point of tangency (x,y) our slope of tangency is just -x/y (and slope of normalcy is y/x)

  84. anonymous
    • one year ago
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    ok

  85. amistre64
    • one year ago
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    we then construct our lines using our tangency points and the slopes required. y-b = m(x-a) for tangent points (a,b) and slope m

  86. amistre64
    • one year ago
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    for the point (2,3) , tangent slope is -2/3 and normal slope is 3/2 lines are: tan y = -2/3 (x-2)+3 norm y = 3/2 (x-2) +3

  87. amistre64
    • one year ago
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    it just occured to me that the normal line is going thru the origin anyways so y = 3/2 x is what it simplifies to :)

  88. anonymous
    • one year ago
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    ok and the point (-18/5, -1/5).. tangent slope is -18 and normal slope is 1/18 lines are tan y=-18(x+18/5)-1/5 normal y=1/18(x+18/5)-1/5

  89. amistre64
    • one year ago
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    yep, norm y can be simplified as y = 1/18 x the constant zeros out becuase the y intercept is the origin for the normal line.

  90. anonymous
    • one year ago
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    ok cool... one question... did u wrote all this out on paper? i want to see the consistency of each step like without my part... can u attach it ..

  91. amistre64
    • one year ago
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    i did not, i wrote it out in the post and whats left in my head is bound to flitter away int oblivion by tomorrow :)

  92. anonymous
    • one year ago
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    ok thanks ..i will go through back all the post to make sure i understand everything

  93. amistre64
    • one year ago
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    my idea is only one method .... im sure there is a simpler approach. but what i did was: i found that the solution points generalize to the line: y = (-ax+k)/b sub that into the circle to determine x values; which in turn can be used to define y

  94. amistre64
    • one year ago
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    use the points of tangency to define slopes and lines as needed

  95. amistre64
    • one year ago
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    good luck

  96. anonymous
    • one year ago
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    i think my problem here is ..i dont fully understand this tangency thing ...i will watch some videos on it..

  97. anonymous
    • one year ago
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spraguer (Moderator)
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is replying to Can someone tell me what button the professor is hitting...

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