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HibaSyed Group Title

Please I really need help with this one.Find an equation of the circle passing through (6, 7) and tangent to the line y = 2x + 1 at (5, 11). Write your answer in standard form.

  • one year ago
  • one year ago

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  1. anneflys Group Title
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    So to find the tangent line, you have to derive. You understand this part right?

    • one year ago
  2. anneflys Group Title
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    And then to derive...y'=2

    • one year ago
  3. anneflys Group Title
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    When you derive, you get the slope...So the slope of your tangent line is 2.

    • one year ago
  4. anneflys Group Title
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    So now you put this into point slope form y-7=2(x-6)

    • one year ago
  5. anneflys Group Title
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    And then to put it into standard form you just do some algebra: y=2x-12+7-->2x-5

    • one year ago
  6. HibaSyed Group Title
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    what would I get

    • one year ago
  7. anneflys Group Title
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    y=2x-5

    • one year ago
  8. cinar Group Title
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    |dw:1355017036124:dw|

    • one year ago
  9. cinar Group Title
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    x^2+y^2=r^2 sub. x=6 y=7 find r=sqrt85

    • one year ago
  10. cinar Group Title
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    you should find coord. of center of circle..

    • one year ago
  11. cinar Group Title
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    you can use point-line distance formula to find coordinate of center..

    • one year ago
  12. cinar Group Title
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    http://mathworld.wolfram.com/Point-LineDistance2-Dimensional.html

    • one year ago
  13. HibaSyed Group Title
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    So I just plug Xand Y into the equation ax+by+c=0 but how would I find a,b,c

    • one year ago
  14. cinar Group Title
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    y = 2x + 1 => 2x-y+1=0 a=2, b=-1, c=1

    • one year ago
  15. HibaSyed Group Title
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    Right, and what would I do after that how, do I convert it in standard from for a circle

    • one year ago
  16. cinar Group Title
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    (x-x1)^2+(y-y1)^2=r^2 where x1 and x2 coordinate of center..

    • one year ago
  17. HibaSyed Group Title
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    and x1 and y1 would be 6, and 7

    • one year ago
  18. cinar Group Title
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    no, you need to find it..

    • one year ago
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