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anonymous

  • one year ago

By finding the points of intersection, show that the line through (0,5) and (3,9) is tangent to the circle x*2+y^2=9

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  1. anonymous
    • one year ago
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    you got the equation for the line?

  2. anonymous
    • one year ago
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    thats all the question give

  3. anonymous
    • one year ago
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    ok let me rephrase the question can you find the equation for the line?

  4. anonymous
    • one year ago
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    |dw:1443226608217:dw|

  5. anonymous
    • one year ago
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    ??

  6. anonymous
    • one year ago
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    yeah good

  7. anonymous
    • one year ago
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    so now to find the point of intersection replace \(y\) in \(x^2+y^2=9\) by \(\frac{4}{3}x+5\) and solve \[x^2+(\frac{4}{3}x+5)^2=9\] and see that there is only one solution to that quadratic

  8. anonymous
    • one year ago
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    x^2+(16/9)x^2+(40/3)x+25=9 (25/9)x^2 + (40/3)x=-16

  9. anonymous
    • one year ago
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    should get \[(25 x^2)/9+(40 x)/3+25 = 9\]

  10. anonymous
    • one year ago
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    oh yeah, what you got

  11. anonymous
    • one year ago
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    multiply both sides by 9

  12. anonymous
    • one year ago
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    \[25x^2+120x+144x=0\] which , by some miracle , is a perfect square

  13. anonymous
    • one year ago
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    \[(5x+12)^2=0\\ x=-\frac{12}{5}\] is the only solution

  14. anonymous
    • one year ago
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    typo there, i meant \[25x^2+120x+144=0\]

  15. anonymous
    • one year ago
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    ok so thats how we show that the line through the given points is tangent to the circle?

  16. anonymous
    • one year ago
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    x^2+y^2=9 -12/5+y^2=9 y^2=57/5 y=sq.rt (285)/5

  17. anonymous
    • one year ago
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    @satellite73

  18. anonymous
    • one year ago
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    this\[25x^2+120x+144=0\]s the quadratic equation you get

  19. anonymous
    • one year ago
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    the fact that it has one solution (not 2, not 0) is what tells you that the line is tangent to the graph

  20. anonymous
    • one year ago
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    ok and if it had 2 solutions then its not tangent to the graph?

  21. anonymous
    • one year ago
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    yes it would cut the circle twice

  22. anonymous
    • one year ago
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    and if you had no solutions, it would not cross the circle at all

  23. anonymous
    • one year ago
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    ok

  24. anonymous
    • one year ago
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    thank you

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