anonymous
  • anonymous
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
Mathematics
schrodinger
  • schrodinger
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anonymous
  • anonymous
Use completing the square to get them both in the form of :\[ (x-h)^2+(y-k)^2=r^2 \]
anonymous
  • anonymous
For example: \[ x^2-4x = (x-2)^2-4 \]
anonymous
  • anonymous
i still cant work it wio

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anonymous
  • anonymous
General equation: x^2 + y^2 +2gx + 2hy + c=0 Centre ( -g,-f)
anonymous
  • anonymous
x^2+y^2-4x-2y-11=0 Centre (2,1)
anonymous
  • anonymous
x^2+y^2+20x-12y+72=0 Centre (-10,6)
anonymous
  • anonymous
Hope this will help u..
anonymous
  • anonymous
Still gotta find radius though. @Yahoo!
anonymous
  • anonymous
\[r=\sqrt{g^2+f^2-c}\]
anonymous
  • anonymous
You really learn that in class?
anonymous
  • anonymous
@yahoo how do u get the center x^2+y^2+20x-12y+72=0 Centre (-10,6)?
anonymous
  • anonymous
Compare it with the General equation..
anonymous
  • anonymous
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)
anonymous
  • anonymous
@lissafi Do you know how to complete the square?
anonymous
  • anonymous
not on this
anonymous
  • anonymous
Just ignore the \(y\) terms and try to complete the square for (x\), then ignore the \(x\) terms and complete the square for \(y\)
anonymous
  • anonymous
wio is that what yahoo did?cos i am lost on this method
anonymous
  • anonymous
I dunno what Yahoo! did...
anonymous
  • anonymous
\[ x^2+y^2-4x-2y-11=0 \]Okay so we just ignore the \(y\) terms... \[ x^2-4x-11=0 \]How do you complete the square here?
anonymous
  • anonymous
wio i dont know much on completing the square
anonymous
  • anonymous
Look it up...
anonymous
  • anonymous
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
anonymous
  • anonymous
@lissafi u got it..now
anonymous
  • anonymous
yap,and for x^2+y^2+20x-12y+72=0?
anonymous
  • anonymous
Try it out...
anonymous
  • anonymous
i lost in (x-2)^2 + (y-1)^2 = 16
anonymous
  • anonymous
i know you add +4+1 on both side
anonymous
  • anonymous
Since we added 4 and 1 on LHS we have to do in RHS tooooo
anonymous
  • anonymous
i know you can find x to subtitute to get y and have the points,but i dont know this method
sirm3d
  • sirm3d
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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