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lissafi

  • 3 years ago

show that the two circlex^2+y^2-4x-2y-11=0 and x^2+y^2+20x-12y+72=0 do not intersect.Hint;find the distance between their centers

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  1. wio
    • 3 years ago
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    Use completing the square to get them both in the form of :\[ (x-h)^2+(y-k)^2=r^2 \]

  2. wio
    • 3 years ago
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    For example: \[ x^2-4x = (x-2)^2-4 \]

  3. lissafi
    • 3 years ago
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    i still cant work it wio

  4. Yahoo!
    • 3 years ago
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    General equation: x^2 + y^2 +2gx + 2hy + c=0 Centre ( -g,-f)

  5. Yahoo!
    • 3 years ago
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    x^2+y^2-4x-2y-11=0 Centre (2,1)

  6. Yahoo!
    • 3 years ago
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    x^2+y^2+20x-12y+72=0 Centre (-10,6)

  7. Yahoo!
    • 3 years ago
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    Hope this will help u..

  8. wio
    • 3 years ago
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    Still gotta find radius though. @Yahoo!

  9. Yahoo!
    • 3 years ago
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    \[r=\sqrt{g^2+f^2-c}\]

  10. wio
    • 3 years ago
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    You really learn that in class?

  11. lissafi
    • 3 years ago
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    @yahoo how do u get the center x^2+y^2+20x-12y+72=0 Centre (-10,6)?

  12. Yahoo!
    • 3 years ago
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    Compare it with the General equation..

  13. Yahoo!
    • 3 years ago
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    x^2 + y^2 + 2gx + 2fy + c=0 C ( -g,-f) so Here x^2+y^2+20x-12y+72=0 2g = 20 g = 10 2f = -12 f = -6 Since the Centre is -g,-f C ( -10 , - (-6) ) C ( -10 , 6)

  14. wio
    • 3 years ago
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    @lissafi Do you know how to complete the square?

  15. lissafi
    • 3 years ago
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    not on this

  16. wio
    • 3 years ago
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    Just ignore the \(y\) terms and try to complete the square for (x\), then ignore the \(x\) terms and complete the square for \(y\)

  17. lissafi
    • 3 years ago
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    wio is that what yahoo did?cos i am lost on this method

  18. wio
    • 3 years ago
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    I dunno what Yahoo! did...

  19. wio
    • 3 years ago
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    \[ x^2+y^2-4x-2y-11=0 \]Okay so we just ignore the \(y\) terms... \[ x^2-4x-11=0 \]How do you complete the square here?

  20. lissafi
    • 3 years ago
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    wio i dont know much on completing the square

  21. wio
    • 3 years ago
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    Look it up...

  22. Yahoo!
    • 3 years ago
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    x^2+y^2-4x-2y-11=0 x^2-4x + y^2 -2y = 11 x^2 - 4x + 4 + y^2 - 2y + 1 = 11 + 4 + 1 (x-2)^2 + (y-1)^2 = 16 (x-2)^2 + (y-1)^2 = 4^2 now it is a circle with centre (2,1) and radius 4

  23. Yahoo!
    • 3 years ago
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    @lissafi u got it..now

  24. lissafi
    • 3 years ago
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    yap,and for x^2+y^2+20x-12y+72=0?

  25. Yahoo!
    • 3 years ago
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    Try it out...

  26. lissafi
    • 3 years ago
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    i lost in (x-2)^2 + (y-1)^2 = 16

  27. lissafi
    • 3 years ago
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    i know you add +4+1 on both side

  28. Yahoo!
    • 3 years ago
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    Since we added 4 and 1 on LHS we have to do in RHS tooooo

  29. lissafi
    • 3 years ago
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    i know you can find x to subtitute to get y and have the points,but i dont know this method

  30. sirm3d
    • 3 years ago
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    in \[x^2+y^2+20x-12y+72=0\] what term must be added to \[x^2+20x\] to get a perfect square trinomial that can be written in the form \[(x+\_\_)^2\]

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